Helping niece. Getting very high numbers. Reddit Pythagorean thm & quadratic equation r/homeworkhelp
Вставка
- Опубліковано 28 чер 2024
- We have a right triangle with lengths x, x+6, and 30. We have to use the Pythagorean theorem to set up an equation to solve for x. This results in a quadratic equation with high numbers that we have to solve, which we can do by either factoring or by using the quadratic formula. Subscribe to @bprpmathbasics for more math tutorials.
This question is from Reddit r/Homeworkhelp. See the original post here / 9jhwwjxg28
0:00 Solving for x from a right triangle
3:05 Solving the quadratic equation by factoring
8:14 Solving the quadratic equation by using the quadratic formula
Shop my math t-shirts & hoodies on Amazon: 👉 amzn.to/3qBeuw6
-----------------------------
I help students master the basics of math. You can show your support and help me create even better content by becoming a patron on Patreon 👉 / blackpenredpen . Every bit of support means the world to me and motivates me to keep bringing you the best math lessons out there! Thank you!
-----------------------------
#mathbasics #geometry #quadraticequation #quadraticformula
Solving x^2+6x-432=0 by using the pq formula: ua-cam.com/video/NlX_VR-e8qo/v-deo.html
Feels like you failed OP's question: "how to solve it without getting any large numbers". 432 is still 'large'.
@@feha92that was only the uncles complaint , like he was assuming it should give smaller numbers only...
Is there any numerical method available to calculate gamma function of non analytical numbers like (1/3), (1/5) ?
I assumed that this was a multiple of a 3-4-5 triangle. Since 30=6*5 the the other sides would be 6*3=18 and 6*4=24. So x=18 works. Done. LOL. The only issue is that this does not prove uniqueness.
I fully expected him to tackle it being a scaled up 3-4-5. In g10, it’s usually that or a 5-12-13.
He probably just didn't cover that because it's not the right learning moment for it. At this unit in 10th grade they need to practice solving quadratics, doing a tangent into the Pythagorean triples would probably not be super helpful.
You don’t need to know that it is a 3-4-5 to realize that resetting x=6*y will give you a much easier quadratic eqn. When you do that though it’s obvious that it is a 3-4-5
@beefchalupa I agree. Someone experienced in such problems would solve by inspection but you need to get experience. I would have preferred solving the quadratic by completing the square but, again, it takes experience to know that 21^2=441.
@@robertpearce8394 It's pretty easy to figure out that 441 is 21^2, even not knowing it off the top of your head. Consider how the discriminant got to be 441 in the first place. In my solution, my quadratic became the following, after dividing everyone by 2: x^2 + 6*x + 18 - 450 = 0.
The discriminant becomes: 36 - (18 - 450), which all consists of easily recognizable multiples of 9. Thus, 441 is a multiple of 9. Divide by 9 to find out which multiple of 9, and you see that it is 9*49. This tells you that its square root is 21.
I didn't bother factoring, went straight to my boy, the quadratic formula.
I completed the square.
Rewrote 900=2x^2+12x+36 as 450=x^2+6x+18, then moved over the 18 to get 432=x^2+6x. Then completed the square for 432+9=(x+3)^2, which leads to taking the square root of 441 and subtracting 3 from it. 441=9*49, so sqr is 21=3*7, minus 3 is 18.
yeah i didn't feel like factoring manually, love having something you can just plug stuff in and get the right stuff out lol, then you can ignore the negative factor since it's a standard geometric shape.
I'm glad bprp showed both methods though for 9th/10th grade students.
"This problem deceived me so hard, am I stupid?" See the video here: ua-cam.com/video/jCIfoKaHBX4/v-deo.html
With hindsight, knowing the solution, one might have spotted the solution right away.
This is just the most basic Pythagorean triplet 3, 4, 5 all multiplied by a factor 6.
If we divide all sides by 6 and replace x/6 by z, the triangle side lengths present itself as 5, y+1, y, which is the well-known Pythagorean triplet for y=3.
Multiplying with 6 gives us the value of x=18.
Like I said, it looks almost obvious, but that's often the case with the wisdom of hindsight! 😁
And obviously, using this kind of trick is neat but only if you already know the method that works in all cases, not just a very specific one.
Decided to have a go at the math before watching. FOIL does indeed give the numbers OP listed. So next step is factoring 864 keeping in mind that one factor being negative and the other positive, one of them being doubled before they are combined, you are left with positive 12. I found that to be the case with 18 and 48, 18 doubled being 36, subtracted from 48, and leaving 12. So x is equal to 18, -24, or both. Putting them through the original quadratic equation, it turns out they both work, just putting the values of 324 and 576 into the formula in reverse with the negative value as x. If the problem presupposes a positive value, that makes -24 impossible, but if it doesn't, you can represent the correct answer/s as two proportional triangles on opposite sides of the origin on a line graph, one with positive values for both x and y, and the other negative for both but with the longer side on the other axis.
Exactly. You can define -24 as -24 - 0 and -18 as -24 + 6. Both will give the hypotenuse as 30 which will be (-24)^2 + (-18)^2.
The negative solution has a geometric interpretation too -- it's a flipped-over triangle with the horizontal side going 24 to the left and the vertical side going 18 down. So it's just the same triangle in a different orientation.
He know bro, he is a college professor of math. He was simplefing the problem for the 10th grader
That kind of reminds me of some standardized tests where there were explicit warnings about figures not being as drawn and to only pay attention to the numbers, angle marks, and length marks. You could argue to the scorer negative solution works in that setting (if the triangle isn't positioned on a coordinate plane) by saying the makers drew it wrong. 🤣
It's always "What is x" but never "How is x"
We will never know x's true mental health :(
Yes, y broke up with him very recently and he hasn't gotten himself together yet
Why is X!
N is sad too everyone forget him 😢
Thank you, Drax.
I went for completing the square to find x. Got 2x^2 + 12x + 36 = 900, divided away 2 on both sides to get x^2 + 6x + 18 = 450, subtracted 9 on both sides to get x^2 + 6x + 9 = 441 >> (x + 3)^2 = 441 >> x + 3 = ±21 >> x = 18 or x = -24.
genius
To find the sqrt of 1764 without a calculator, I would recommend first seeing if it's divisible by 4 or 9. 1764 is divisible by both, leaving 49. So the sqrt of (4*9*49)=(2*3*7)=42.
I don't see why you just don't complete the square when you get to
x² + 6x − 432 = 0
Bring over the constant 432 to the right hand side and add (6/2)² = 9 to both sides and we have
(x + 3)² = 441
and since 441 = 21² and since x needs to be positive this gives
x + 3 = 21
and so
x = 18
Nicely done!
Very classical reduction. 30=6x5 and you got x+6. So let x=6k , equation with big numbers reduces in a second to k^2+(k+1)^2=25.
k=3 is the positive trivial solution and thus x=6x3=18
Just a thought: if you want to find factors of 432 and the sum is something low (like 6), you could look around the square root for factors. The square root of 432 is about 21 (rounded, in this case a fine place to start. To get a difference of 6, go up 3 and down 3 from that. This is not a perfect solution, but it will often give you a place to start looking!
Since the constant term is negative, we need two numbers that multiply to make 432, and SUBTRACT to make 6.
The two numbers will come from the prime factorization of 432 = 2 x 2 x 2 x 2 x 3 x 3 x 3. We might try: 2 x 2 x 2 x 2 =16 and 3 x 3 x 3 = 27, and get a difference of 11, which is too big (our target is 6). To make the 1st number bigger & the 2nd number smaller, we exchange a 2 in the 1st with a 3 from the 2nd, and get two other numbers: 2 x 2 x 2 x 3 =24 and 2 x 3 x 3 = 18, whose difference is 6. [Use the information from each guess to change the numbers to get closer to the target. Sometimes you get lucky and hit the target in 1-2 tries. Other times, it take 3-5 if you don't start out not that far from the target on the first try...]
Lastly, because the sign of the target is positive, we want the larger number to carry the positive sign; that is, the two.signed-numbers are: +24 and -18, which gives the factorization of the polynomial:
x² + 6x - 432 = (x + 24)(x - 18)
You can take the trial and error out of this as follows. To factor the left hand side of
x² + 6x − 432 = 0
we are looking for two numbers with sum 6 and product −432. Since the sum of the numbers we want to find is 6, their _average_ is half of this, which is 3. Now, the two numbers we need to find are obviously at equal distances from their own average. So, if _h_ is _half their difference_ the smaller of the two numbers (assuming h to be positive) is 3 − h and the larger of the two numbers is 3 + h. And since their product must be −432 we have
(3 − h)(3 + h) = −432
Applying the difference of two squares identity (a − b)(a + b) = a² − b² this gives
9 − h² = −432
−h² = −441
h² = 441
h = 21
So, the numbers we were looking for are 3 − 21 = −18 and 3 + 21 = 24 and the quadratic therefore factors as
(x − 18)(x + 24) = 0
giving the solutions x = 18 and x = −24.
incredible factoring. thanks
I noticed that each integer in the question is a multiple of 6. if you do a quick substitution of x = 6y, then all side of the triangle is a multiple of 6. we can then reduce the problem using a similar triangle with sides {y, y+1,5} and solve for y.
y^2 + (y+1)^2 = 5^2
2y^2 +2y -24 = 0
y^2 +y -12 = 0
(y-3)(y+4) = 0
y=3 or -4
x = 18 or -24
a lot easier than working with 12 than 864
There are only 5 pythagorean triples with a hypotenuse of 30 or less, but each triple has at least one side with a prime length, therefore the only solution (if it exists), must be a multiple of (3,4,5). 30 = 5 times 6, therefore x=18 and (x+6)=24,
Seeing the 30, my first thought it was a scaled up 3,4,5 triangle, and it is....
The large numbers you can get for the c in the a,b,c-formula is why I don't like it. I very much prefer to use the p,q-formula which is much easier to remember since there are only two variables, because in order to use it, you have to divide by the factor in front of x², which is something that you also did in this case.
x = -p/2 ± √((p/2)² - q)
No need to multiply a big number c by 4a in the square root.
I prefer the m-n-formula, which I made up myself... ;) You just have to bring the quadratic equation into the form 0.5x² + mx + n = 0, then the solutions are simply x1,2 = -m ± √m² - 2n.
@@bjornfeuerbacher5514
Well, the p,q-formula is how I learned it in school. The a,b,c-formula is also taught in schools, but I don't like it.
Maybe if you lobby for your new formula, it too will be taught in schools around the world.
@@Matty0311MMS I'll try. :D
@@bjornfeuerbacher5514 why the (1/2)x² term though? 3Blue1Brown showed the m ± sqrt(m²-p) formula which is equivalent to the pq-Formula but it appears somewhat simpler than yours to me, because I don't understand why you would use the 0.5 coefficient
@@jellymath Look up the meaning of m and p in 3Blue1Brown formula, then you'll see that it's equivalent to mine.
I choose 1/2 in my formula in order to simplify the usual a-b-c formula, which has 2a in the denominator. For a = 1/2, the denominator then simply is 1.
There are a bunch of ways to solve this. But there is a clever secret hidden after the FOIL. Leave 30^2 as is and bring it to the left side. You’ll get 2x^2 + 12x + (36 - 30^2). You get a lovely difference of two squares which makes solving near trivial and no calculator is needed! Brilliant!! For completeness:
2x^2 + 12x + (6^2 x- 30^2)
= 2x^2 +12x + (6-30)(6+30)
= 2x^2 +12x + (-24)(36)
= x^2 + 6x - 24*18 (if you divide 24 by 2 instead of 36 you don’t get factors that differ by 6)
= (x + 24)(x - 18)
This then gives you the solution as presented but you’ve done it completely by hand😊
Nice solution.
Small typo. You have an extra x , in "2x^2 + 12x + (6^2 x- 30^2) "
should be "2x^2 + 12x + (6^2 - 30^2) ".
"I can't imagine the teacher giving her something this complex!"
The problem: use the Pythagorean theorem then solve a quadratic equation. 🙄
Don't even need to solve the quadratic equation, completing the square is much easier here in terms of calculations, no big numbers there.
But but big number!
Just wait until answers to this kid’s math problems like this start involving square roots or literal complex numbers, their parents are gonna freak out
Me: **looks at it for a few seconds** "18."
I think the main problem here was just the initial assumption that "large numbers = hard problem". The size of the numbers doesn't change the problem or how to solve it. If you do the right steps, you'll get the right result, no matter what the actual numbers are.
basically this.
1 + 1 = 2 is no less complicated than 1000 + 1000 = 2000
The underlying principle remains the same and the method to solve either is as well.
As others already posted, it avoids a lot of algebra to recall that 3,4,5 is a Pythagorean triple. Since 30 is on the hypotenuse, we can divide all sides by 6 to give 30/6, x/6, x/6+1. That would make x/6 = 3 and x = 18. The other side would be x+6 = 18+6 = 24, and 24/6 = 4, so it checks out that we have 3,4,5 proportions, and the problem is solved. This should be tried first because school homework problems always tend to have "easy" solutions and we can avoid complicated methods unless we are forced into it.
Somehow this video popped up in my recommendations. Couldn't find any mention for other approach, so let me expain it:
x² + (x+6)² = 30²
x² = 30² - (x+6)²
Using formula a² - b² = (a-b)(a+b):
x² = (24-x)(36+x)
x² = -x² - 12x + 24*36
The point here is not to multiply numbers before needed.
2x² + 12x - 24*36 = 0
x² + 6x - 24*18 = 0
Now using: (_1, _2 are indexes)
{ x_1 + x_2 = -6
{ x_1*x_2 = -24*18
No need to perform calculations, roots are already there:
x_1 = -24 < 0
x_2 = 18
Only "serious" calculation we did was 24 - 36 when transforming (24-x)(36+x), so this is totally solvable even without writing.
Another point is that when we get to
x² + 6x - 432 = 0
You can use not b²-4ac, but rather k²-ac (because b is dividable by 2), where k=b/2. Roots are calculated as follows: (-k ± √(k²-ac))/a
In this case you will get only 441 under sqrt, which is only 21²
At the very start set 6u=x
then scale the entire triangle down by a factor of 6
by pythagoras you get u² + u² + 2u + 1 = 5²
2u² + 2u - 24 = 0
u² + u - 12 = 0
(u+4)(u-3) = 0
since we're looking for a length reject the negative sol, so u=3
finally we know x=6u so x=6*3=18
U can also use vjeta theorem kinda same as the 1st one
couldn’t you also complete the square for this question too? just curious
Yes you could, would give something similar to what quadratic formula gives
@@TrimutiusTooyes thanks
You can solve every quadratic equation if you complete the square
x² + 6x - 432 = x² + 6x + 3² - 3² - 432 = (x + 3)² - 441 = (x+3)² - 21². Probably that's even the fastest and easiest way to solve this...
Sure. It would have been even better just to leave the constant term at the right hand side because then you get
x² + 6x = 432
Add (6/2)² = 9 to both sides and we have
(x + 3)² = 441
and since 441 = 21² and since x needs to be positive this gives
x + 3 = 21
and so
x = 18
No big numbers, no messing with the quadratic formula, no guesswork to find a factorization. The only thing here you need to know is that 441 = 21² but that isn't too hard since 20² = 400.
once you have the quadratic I think the easiest way to deal with it is by completing the square and then just solving for x, only hard bit might be taking sqrt 441 but it isn't super hard
Suppose x = 6y
6y + 6 + 6y = 30
Factor
6(y + 1 + y) = 30
Divide everything by 6
y + 1 + y = 5
2y + 1 = 5
2y = 6
y = 3
If x = 6y and y = 3, then x = 6 ⋅ 3
x = 18
If the other side is x + 6, then it is 18 + 6 = 24
The sides are 18, 24, and 30.
Never do maths again
x^2 + 6x - 432 = 0
Note that when you use the quadratic formula, you’re going to have sqrt(6^2 + 4*432). See if you can factor 432 to pull out a 6^2 to make the number smaller.
432 = 2 * 216
= 2 * 2 * 108
= 2 * 2 * 2 * 54
= 2 * 2 * 2 * 2 * 27
= 2 * 2 * 2 * 2 * 3 * 3 * 3
= 12 * 6^2
Now you have
-6/2 +- sqrt( 6^2 + 4*12*6^2 )/2
Factor out the 6^2 outside the sqrt and divide by 2
= -3 +- 3sqrt( 1 + 48 )
Factor out the 3s, simplify the sqrt
= 3(-1 +- 7)
so x = 3( 7-1) = 18 or x = 3(-1 - 7) = -24.
Take positive solution for it is a real shape.
If we think of the left vertex containing the hypotenuse and the base, and take it as the origin, x=-24 would mean 24 toward the left, and x+6= -24+6=-18 would mean 18 units down the origin, so basically a flipped triangle.
here’s another method to factor x^2 + 6x - 432 = 0, basically you want to split the middle term into two parts such that you can get two factors by grouping and then taking out a common factor
here’s how you do it : for any polynomial ax^2 + bx + c, you want two numbers m and n, such that m + n = b, and mn = ac
then your polynomial will be much easier to factorise
take this polynomial, x^2 + 6x - 432; we want two numbers, m and n, such that :
m + n = 6, mn = -432
solve this equation either by substitution (or by forcing an elimination, which is what i do)
and eventually you’ll get m = 24, n = -18
the polynomial can now be rewritten as x^2 + 24x - 18x - 432
or x(x + 24) - 18(x + 24) and lastly (x + 24)(x - 18)
and yes, the cross method may be more intuitive, but i think this method is also worth a mention
or wait just complete the square 💀
x^2 + 6x - 432 = 0
so x^2 + 6x = 432
add 9 to both sides and
(x + 3)^2 = 441 or 21^2,
x + 3 = 21 (-21 is invalid cus geometry)
and x = 18 💀
much easier wth
@@ve4rexetaking it one step further, you don't need to ever square out the 30.
You get 30²/2 -9 on the right side.
Then you can reformulate it as:
10²3²/2 - 3²
That then leaves you with:
(50-1)3²= 49*3²
Which is easy to take a squareroot of :).
Atleast if you're an incompetent squareroot taker like me
It's a math homework problem. 70% of them are 3-4-5 triangles. Look at the numbers, and you immediately spot that the short sides are x and x+6, and the hypotenuse is a multiple of 6. So take out the factor of 6, and you have a hypotenuse of 6*5, and sides of 6y and 6(y+1), where we've written y = x/6. Oh look - it's a 3-4-5 triangle. y=3, so x = 18.
For the -6 ± √(6)² + 4(432), all over 2 step, i would have taken that 6² and turned it into 2²•3², taken the 4 and turned it into 2², and factored the 2² out of the square root, giving me
-6 ± 2√3²+432 all over 2,
Then i would have factored out the 2 from the numerator and canceled it with the denominator giving
-3 ± √3²+432, becoming -3 ± √441
Prevents you from having to work with 4 digit numbers!
The divisibility by 6 suggests the substitution x=6t.
(6*5)^2=(6t)^2+(6[t+1])^2
The 6's square, factor, and cancel.
5^2=t^2+(t+1)^2
The positive solution t=3 exists, and the RHS is increasing when t>0, so is one-to-one for positive t, so the other solution is negative.
If you don't like big numbers start by scaling the triangle down by the common factor of 6 and setting x=6y. The triangle is then y, y+1, 5 and most of you will spot the answer already.
Nevertheless, we can use Pythagoras to get y^2 + (y+1)^2 = 25. So 2y^2 + 2y - 24 = 0. Divide through by 2 and we have y^2 + y - 12 = 0. To factorise it, we're looking for two factors of 12 that differ by 1. That gives us (y - 3)(y + 4) = 0 meaning y = 3 or -4.
We don't have negative sides, so y = 3. And the solution to the original is then x = 18. No more big numbers and we don't need the quadratic formula.
are you bprp math because you do maths or you do maths because you are bprd math ?
man, I love these quotes.
What is bprp to you? The strongest
Nah, I'd calculate
stand proud, you are strong
bprp stands for black pen red pen, which references him to having a black pen and a red pen in one hand.
Hey, I think this is more like the intended solution, although I don’t think it’s at all obvious especially for a grade 10 student. Apologies for the poor setting, it’s a lot neater to see on paper so it might be worth writing out.
Let u=x+3
Then
30^2=(u-3)^2+(u+3)^2
After distribution, middle terms of each square cancel:
30^2=2(u^2 +3^2)
Rearranging:
(1/2)(3^2)(10^2)-3^2=u^2
Factoring 3^2 out of 30^2:
3^2(1/2*10^2-1)=u^2
u^2=(3^2)(7^2)
u=(+/-)21
x=-3(+/-)21
Exclude negative solution to give x=18.
By normalizing the quadratic form, the numbers become quiet small: 441 under the root which can easily be deduced to be 21^2 (even if you use approximation). The students have to get used to the nature of quadratics doubling the number of digits. There's nothing evil about that. :D
I would suggest to use substitution to reduce the numbers we have to deal with; with y := x/gcd(6,30) >= 0 (or 6y = x >= 0) you get:
x^2 + (x+6)^2 = 30^2
(6y)^2 + (6y+6)^2 = (6*5)^2
y^2 + (y+1)^2 = 5^2
y^2 + y^2 + 2y +1 = 25
2 y^2 + 2y - 24 = 0
y^2 + y - 12 = 0
(y+4) (y-3) = 0
y-3 = 0
y = 3
x/6 = 3
x = 18
Divide the full equation by 2 and you get x^2 + 6x - 432 = 0/2 = 0. From there ypu notice 432 is even and has a digital root of 9, so its a multiple of 18. Hence that's the first number I tried, and -432/-18 = 24 and 24-18=6, so equation solved straight away.
Let u=x/6. Now we have a = 6u, b= 6u+6 = 6(u+1), and c = 30 = 6·5. We can scale the triangle to get rid of the sixes, giving us a right triangle with sides u, u+1, and 5 as the hypothenuse. That one is clearly the 3-4-5 triangle, though you can still work it out now the numbers are not big anymore. Rejecting the solution u = -4, we are left with u = 3, and therefore x = 18.
For this sort of quadratic formula use where you expect integer answers, I prefer to make a=1/2 if possible, meaning the formula effectively reduces to:
x = -b +- sqrt(b^2 - 2c)
so in this case with b = 3 and c = -216:
x = -3 +- sqrt(9 + 432)
x = -3 +- sqrt(441)
x = -3 +- 21
x = -24 or 18
However, for this problem, I actually mentally divided all the lengths in the triangle by 6 and substituted y = x/6, giving the new lengths of the triangle being y, y+1 and 5. This is obviously a 3,4,5 triangle so the original x must be 3*6 = 18
I'm not sure why the OP was so afraid of the big numbers. By grade 10, you're expected to be able to multiply by hand. Even for the square root, I was taught how to compute square roots by hand in grade 9.
It's not very complex, but I will say it seems excessively tedious to do by hand if you're just practicing how to calculate the sides of a right triangle. I skipped two years in math without struggling and this problem still would've raised red flags to me just because the numbers aren't so convenient when they normally are at this grade level.
@@Salien1999all things considered the number is quite convinient, you got integers as roots
@@CyberFlare-fn9kn yeah, it's not TERRIBLE but when put into the context of a 10th grade assignment it's tedious enough to make you wonder if you're doing it the most efficient way possible. I'm sure that's what led to the post.
I would make x = 6y. Scale down the triangle by 6 and solve y²+(y+1)²=5² then substitute back to x
Honestly the expected number is so small my go to would to be brute force the equation until I got the right answer
even though i'm good at maths and saw the solution right away i want to say thank you because i think this was explained well (however in this case when you showed both methods for the quadratic i'd say use the formula because the numbers are too big for factorizeing to be good)
You can take the trial and error out of factoring this equation as follows. To factor the left hand side of
x² + 6x − 432 = 0
we are looking for two numbers with sum 6 and product −432. Since the sum of the numbers we want to find is 6, their average is half of this, which is 3. Now, the two numbers we need to find are obviously at equal distances from their own average. So, if _h_ is half their difference the smaller of the two numbers (assuming h to be positive) is 3 − h and the larger of the two numbers is 3 + h. And since their product must be −432 we have
(3 − h)(3 + h) = −432
Applying the difference of two squares identity (a − b)(a + b) = a² − b² this gives
9 − h² = −432
−h² = −441
h² = 441
h = 21
So, the numbers we were looking for are 3 − 21 = −18 and 3 + 21 = 24 and the quadratic therefore factors as
(x − 18)(x + 24) = 0
giving the solutions x = 18 and x = −24.
Of course it is even easier to factor (and solve) this quadratic by completing the square, as follows.
x² + 6x − 432 = 0
x² + 6x + 9 − 9 − 432 = 0
(x + 3)² − 441 = 0
(x + 3)² − 21² = 0
(x + 3 − 21)(x + 3 + 21) = 0
(x − 18)(x + 24) = 0
x = 18 ⋁ x = −24
@@NadiehFan it's much easier in my opinion to write it as (x+3)²=441
x+3=±21
x=21-3=18
Or x=-21-3=-24
All this with a difference of 2 squares seems overcomplicated
@@grassytramtracksI just wanted to demonstrate two different ways to take the trial and error out of _factoring_ a quadratic into two linear factors. As you demonstrate, this isn't quite as easy as just _solving_ the quadratic by completing the square, but my aim here was a little different. I've also posted other comments to this video where I show, just like you do, that solving the quadratic by completing the square is in fact much easier and faster than _both_ methods shown in the video.
Now of course since you could use a calculator sqrt(1764) was't a problem but it's still usefull to know how to deal with square roots of large numbers:
1)You could distribute the /2 so 3±(sqrt(1764))/2, you can get the 2 under the square root so sqrt(1764/4)=sqrt(441) which is much more managable
2)The special calculation method of rational square roots (there's a good mindyiurdesicions video on it)
You take the ones number and compare it with ones numbers of perfect squares. We take the number squaring so we get
1764
I
V
2/8
We cross out the tens and ones and take the closest not exceeding perfect square so in this example 4 (4²=16
i'd just have used something learned in fifth grade: factor it. 1764 = 2 x 882 = 2 x 2 x 441 = 2 x 2 x 3 x 147 = 2 x 2 x 3 x 3 x 49 = 2 x 2 x 3 x 3 x 7 x 7 = 2² x 3² x 7², so in this case nothing remains under the root: sqrt(1764) = 2 x 3 x 7 = 2 x 21 = 42. In emergencies doable without writing anything down 😛
I used the equation method with a calculator just like the video, but in hindsight feel silly for not dividing everything by 6 first!
When attempting factoring, I first start by looking at the sign of c. If positive, the factors will be a sum. If negative, the factors will be a difference.
In this instance, with a negative c and thus a relatively small difference, the factors have to close to the square root of 432. Not knowing that off hand but knowing 20*20 = 400. So with difference a 6 and knowing 432>400 I came up with 18 & 24 first guess.
Is this approach widely taught/used?
When I taught this, I always told my students to be on the lookout for a sneak 3-4-5. So, by inspection . . .
>10th grade
>geo
>gives us a middle school problem
Should I be concerned about our future
2x^2 + 12x + 36 = 900
x^2 + 6x + 18 = 450
(x+3)^2 + 9 = 450
(x+3)^2 = 441
x+3 = +-21
x1 = 18
x2 = -24 (invalid by geometry)
You don't need the big numbers and those complicated solutions.
Can easily do it in your mind in a minute.
What are you doing for step 2?
I am confused why after summarization you are not grouping big numbers by their primes and write 2^4*3^6 or whatever. In this way you xan slowly reduce big numbers. Or just divide by 4 and put the 2 before the square root. do again until you go to 3 and so on. That is the whole reason why I think division rules are so helpful to quickly see if a bigger number is divideable by primes or squares of primes. It is easier for fractions as well as square roots and gives you a feeling for big numbers.
Teachers back in the day: you arent gonna have a calculator to solve all the time
Bprp: just use a calculator for square root
This one isn't too bad without a calculator. If you know long division, you can easily find that 1764=4*9*49, all which are perfect squares.
@@bethanymussman8000 I got that. I'm just making fun of what I was taught when I was a kid. Ironically, every math teacher I had needed a calculator after telling us we weren't gonna use it in college and the such
instead of the quadratic we Always use the p-q-formula in Germany and i honestly think its superior, i got 441 under the root which u can doin ur head to get 21
Looks at problem.
Notes 30 is divisible by 5.
"I think I know where this is going..."
Usually problems are designed to factor out nicely but i when i got x2+whatever+432 i thought i messed up my math...nope just "messy" problem. 3,4,5 wouldve been the way to go but it wasnt allowed and im 25+ year our from geometry didnt even notice it.
The much easier way is to just divide it by 6 first so you get (x/6)^2+((x/6)+1)^2=5^2. Replace (x/6) by a or other variables and voila. Much easier.
six is even so instead of x and x+6 we can have t-3 and t+3, where t=x+3, x=t-3
pithagorean theorem gives us
t^2-6t+9+t^2+6t+9=30^2
2t^2+18=900
t^2=450-9=441 which is the famous 21^2 so t=21 x=21-3=18
much easier than factoring 432
wait how have i never learned this method? lol. Interesting.
Funny, I'm the worst at transposing digits in long numbers and flopping plus and minus, so I guess I was extra careful and solved this in one shot.
Often I need to solve a problem several times and see what value I get the most often. :/ But then I backtrack and prove it has to be the answer.
At my days in pre engineering school. We learned to memorize the results of (a+b)^2 (a+b)'(a-b) and
(a-b)^2.
Do they still do that? It wasn't that important but it could save some time and perhaps avoid slip of the pen
My precal class covered those common cases, though in computer science dealing with numeric analysis and covering cubic spline functions, I wish they would have gone over the basics of cubics more lol
What?!? They said you to memorize those identities in your engineering school?!? Really?? Bruh, whereas me who learnt it four years back that is, in 7th grade lol and still need to use it
30^2 = (x + 6)^2 + x^2
30 = 2x - 6
30 + 6 = 2x
36/2 = x
x = 18
Now at step 2, I made (x+6)^2 + (x)^2 into 2x - 6 and just forgot all the exponents and it works.. Now I feel like this is wrong obviously, could someone please point me in the right direction?
Thanks
use prime factorisation it is easy
Huh grade 10 not able to solve that? Wasnt expecting it, the syllabus here is too different ig?
I am Indian. This was the easiest sum I got so far.
@@user-lb3ex6yh9uyep just wrote a comment abt that, these american dumbasses cant solve that? they would cry if they saw NTSE papers easiest math question💀
grade 10, so she's about 16? Highschool? And the person asking the question think factoring a 3-digit number is "too complex" for the niece's homework. For me, prime factorization is taught at 13, the quadratic formula is taught at 14. I'm shocked that this kind of basic algebra is only taught in highschool.
Anyway, if you're stuck on factoring (especially when c is big), just use the quadratic formula. Factoring method's core is trial and error, and getting the right number fast needs practice, and some luck. You need to factorize a number anyway, so just do it on the big number in the square root and get the answer immediately. (Then pretend you got it by factoring if you want, nobody care how you really got the answer)
In india (10th grade) this problem would probably come for 1 mark, in a test of 80 marks
I thought the lower side was z based on the thumbnail on phone😂 I'd not solve two unknown variables
Solve the integral
integral (√tanx+√cotx)dx
I am pretty sure that's not basic maths
@@user-lb3ex6yh9u its quite interesting and has different approaches to solve
Given:
integral sqrt(tan(x)) + sqrt(cot(x)) dx
Rewrite cot(x) with its definition:
cot(x) = 1/tan(x)
sqrt(cot(x)) = 1/sqrt(tan(x))
Thus:
integral sqrt(tan(x)) + 1/sqrt(tan(x)) dx
Let u = sqrt(tan(x)).
Take derivative:
du = 1/(2*sqrt(tan(x)) * sec(x)^2 dx
Solve for dx:
dx = 2*sqrt(tan(x))/sec(x)^2 dx
dx = 2*u*cos(x)^2 du
Rewrite cos(x) in u-world:
x = arctan(u^2)
cos(arctan(u^2)) = 1/(u^4 + 1)
Construct integral completely in u-world:
integral (u + 1/u) * 2*u/(u^4 + 1) du
Simplify:
integral (2*u^2 + 2)/(u^4 + 1) du
Set up a product of two quadratics to equal the denominator. We know the first and final terms are u^2 and 1, but the middle terms are unknowns, a & b.
(u^2 + a*u + 1)*(u^2 + b*u + 1)
Expand & gather:
u^4 + (a + b)*u^3 + (a*b + 2)*u^2 + (a + b)*u + 1
Set up system to solve for a & b:
(a + b) = 0
a*b + 2 = 0
Solution:
a = sqrt(2), b = -sqrt(2)
Thus:
u^4 + 1 = (u^2 + sqrt(2)*u + 1)*(u^2 - sqrt(2)*u + 1)
Complete the square on both expressions:
u^2 + sqrt(2)*u + 1 = (u + sqrt(2)/2)^2 + 1/2
u^2 - sqrt(2)*u + 1 = (u - sqrt(2)/2)^2 + 1/2
Use partial fractions to rewrite integrand:
(2*u^2 + 2)/(u^4 + 1) = (A*(u+sqrt(2)/2) + B)/((u + sqrt(2)/2)^2 + 1/2) + (C*(u - sqrt(2)/2) + D)/((u - sqrt(2)/2)^2 + 1/2)
Solve for constants:
A = 0, B = 1, C= 0, D = 1
Thus:
(2*u^2 + 2)/(u^4 + 1) = 1/((u + sqrt(2)/2)^2 + 1/2) + 1/((u - sqrt(2)/2)^2 + 1/2)
This is a standard trig sub, in the form of
integral 1/((x + p)^2 + q^2) dx = p*arctan((x + p)/q)
Plugging in problem-specific values for p & q, we get:
sqrt(2)*arctan(sqrt(2)*u + 1) + sqrt(2)*arctan(sqrt(2)*u - 1)
Rewrite in x-world, add +C, and we're finished:
sqrt(2)*arctan(sqrt(2*tan(x)) + 1) + sqrt(2)*arctan(sqrt(2*tan(x)) - 1) + C
Aint no way this is 10th grade. Im in 10th grade rn and were teaching analytical geometry and applying that with vectors, and also random ahh functions.
Why can't you have negative length?
Wouldn't it just go the other way?
Like be a mirror imaged side?
Length does not have a direction. As you walk towards someone and pass them, the distance between you does not become negative; it begins increasing again.
If you bring in directionality, then negative values can exist and start to make more sense, but this starts walking into vectors. Vectors are numbers that have both a direction and a magnitude. The negative of a vector is the same magnitude (length) but in the opposite direction. Even in vectors, though, the start and end of a vector will always have a positive distance between them.
a + b + c = 6
a² + b²+ c² = 14
a³ + b³ + c³ = 36
ab + bc + ac = 11
abc =6
Values of a , b and c are ?
Help me, solve this
I really wonder why bother at all with ax^2+bx+c=0 if you always can use x^2+px+q=0. Esp in this case you get smaller numbers like sqrt(441), which is 21. I really don't get it, in fact, when I was in school, we never ever used a,b,c. Is there any comprehensible advantage using abc-form? I don't see any.
no double fractions ever, which cause a lot of attention errors.
@@emilwandel Double fractions? Where? In this example here, the pq-formula is way faster and easier to calculate.
42? Teacher a Douglas Adams fan?
Well, the 30 and the 6 just scream for dividing everything by 6....
I apologize but that's really really easy question.
Is this what 10th grade is like in the US? I was confused and thought I misunderstood the task, but no, it is just that easy....
Exactly! I don't know what's the program in US but in most post-soviet countries students must know (Not should. Must!) how to solve it at least by the end of the 8-th grade. In some schools even earlier.
I'd expect this to be an Algebra 1 question, which is the "standard" (ie. not accelerated) math class for 9th grade. Kids on the accelerated math track take Algebra 1 in 7th grade. It's quite possible that the poster's child is on a slower, remedial-level track and only encountering algebra in the 10th grade.
@somebodyelse9011 I won't argue that, since I don't know too much about American schools and I was in the "Advanced Science" program in my school, my perception is skewed.
It just seems weird to me: US universities are on pace with the best ones around the world (and that I know 100%), but their schools are generally behind by a lot, it seems.
I think I see why the dropout rates and struggles are so much higher among their students...
3 digit numbers are too big for high school students? I'm a bit worried about our school system.
Why you don't use Delta ? it's the base for 2nd degree equations
Delta = b²-4ac right? its already in the "Baskara formula" but squared already, its a skip step but you calculate the Delta aside and plug in square root of Delta
@@AndreChaosweapon oh ok thanks I didn't know this was a shorcut, but for this level I think using Delta is easier to understand
11:37 we cant you calculator in our region i use a very nice trick,it is called eyeball the closest number and try, so like it must be way greater than 25 or 35 then i just hop to say like 82, last number is 2 cuz you know 2 times 2 gives 4 and the last number is 4
but... take this trick with a grain of salt cuz it is highly ineffective and i am CLEARLY not at smart as this dude who frequently blows my mind away....
If you only take the positive number, why do the whole +- part. I understand the why, but if we know from the start it'll be discarded, why not just streamline it?
Because its not necessarily true that a quadratic always produces a negative result. If you discard the minus right off the start you can miss a valid solution.
It's not good practice anyway because you eventually move on from geometry where everything needs to be positive.
So 3 digit numbers are large now. Given the average American can't count above 12, I guess it's kind of expected.
I mean... this is not really too tough of a question by grade 10 standards, maybe the teacher was just too easy-going
Bro i am grade 6 and i figured it out without help
takes 1 min using pq
American education is too easy. In india this is probably a hard 8th grad question or easy 9th grade question
Your country isn't relevant, no one cares about India
Amerocan schools got fucked by covid. This would be an 8th or 9th grade problem nornally
Could you prove how x^n polynomials always have n number of roots?
You only need to prove it has at least one, that it has n falls from that. The Fundamental Theorem of Algebra is one of the few Fundamental Theorems that can not be fully proven in the class level it is used. The proof I like needs continuity of a function in polar, and the continuity of functions that are near that function but off by a delta theta. A hand wavy proof is shown on sites like ua-cam.com/video/J2iWsiFv8uI/v-deo.html which makes some assumptions, but gives the general outline on how the actual proof I use work. There are proofs of the FT of algebra in other fields, including complex analysis.
That's only true of x is a complex number, and understanding the proof requires a fair amount of knowledge about complex analysis.
@@bjornfeuerbacher5514 You can understand majority of it before complex analysis, but you have to make some assumptions and press the I believe button (especially on the continuity part of the argument).
@@ingiford175 Well, in that case I wouldn't call it an actual proof. ;)
@@bjornfeuerbacher5514 I agree, it is more of a 'proof' in a high school sense, not a proof with rigor. Or a proof in more of a Euclid, where there are some gaps (parallel postulate assumed, intersections of objects assumed)
Yo 345 triangle 😊
quah-chan-tih equation 😅😅😅
Step 1: Scale everything down by 6
x' = x/6
a' = a/6 = x'
b' = b/6 = x'+1
c' = c/6 = 5
Step 2: Plug into Pythagorean theorem and solve for x'
5^2 = (x')^2 + (x'+1)^2
25 = x'^2 + x'^2 + 2x' + 1
25 = 2x'^2 + 2x' + 1
0 = 2x'^2 + 2x' - 24
0 = x'^2 + x' - 12
0 = (x'+4)(x'-3)
x' must be positive, so x' = 3
Step 3: Scale everything back up by 6
x' = 3
x = 6x' = 18
I'm sorry grade 10? 14 years old?
... And a three digit number is too big for them? Is America really...
I was like #500 :D
Non ironycally
While doing in my mind, I found at the first take the couple -18, 24
It's called luck
I've always succeeded at just guessing consecutive pairs and other stuff like it, but that doesn't really matter in a class you still need to show you can do it algebraically. But at least guessing the answer makes creating the work step by step trivial
Me too , but it's wrong
u took 900 as c ig, u should have taken 864 and it would be right!
Easiest way to think of it IMO is we know the original number was a multiple of 3 and 2, but we needed a difference that's 6(a multiple of 3 and 2), meaning both of our factors are multiples of 6. From there it's just a case of seeing what the factors could be.
432 is much too big to be 6,12 or 12,18(20×20=400, so at least one of the numbers must be above 20), so 18,24 is the first plausible answer.
goddamn they dont know this much in 10th? i thought they taught more in america because this is the basics of quadratic equations. here we have NTSE exam which is a big scholarship exam and its really fucking hard. and i dont think thats 10th grade math there, so ig he must wrote the wrong grade because this is too easy
average 10th class queschens
Why are people doing quadratics in Year 10? We didn't learn them until year 11, and only in Maths Methods: the advanced maths class.
We learned it in 9th
18 and 24
Did you watch the video? You are kind of wrong
@@afre3398 I think he meant one side is 18 and the other is 24
If that's so hard, he should come to India 😂
The teacher switched +24 to -24 & -18 to +18 from the value of a & b. $tudents got tricked by the teacher who turned an easy problem into a difficult one.The teacher should have some explaining to do if the whole class failed and force the teacher to switch back solutions the original + 24 & -18. Teachers should be punished for making life difficult for 10 graders. This is not India, China or Korea where they do this and torture and pressure students to commit suicide.