I did it by finding the factors of 92. Much simpler. 92/2 = 46; 46/2 = 23 (prime). Factors are 2, 2, 23. For our problem solution 2x2 = 4. 4x23 = 92. 4 +23 = 27. QED Don't ever make a problem more difficult that it has to be. Solving a quadratic to get this solution is unnecessarily complex.
We have to set up an equation. So X is 'number 1' and 27 - X is 'number 2'. Then you get: X(27 - X) = 92 => 27X - X^2 = 92 => - X^2 + 27X - 92 = 0 => X^2 - 27X + 92 = 0 (everything divided by -1) => (X - 23)(X -4) = 0 => X 1 = 23 and X 2 = 4. So the numbers are 23 and 4 (23 + 4 = 27 and 23 x 4 = 92).
I think in this case, using a quadratic equation for this particular problem is like using a bucket-loader to dig a fence post hole. I set up a table of terms that added up to 27: 1+26 - nope; 2 +25 - nope; 3 + 24-nope; 4+23-bingo! Eight quick steps, one solution. I did assume that it would be a single-digit number, and a double-digit number.
@@Kleermaker1000 great your mind can eliminate steps... in fact I'll observe that maturity in algebra is such. We ALL know to solve x+6 = 2x +8 is -x = 2 Without doing "Subtract 'x' from both sides." "Subtract '6' from both sided." The process of being highly structured helps eliminate mistakes. School math is a combination of "how to" and "how fast." Actual employed mathematicians leans less on speed and more on process and ACCURACY.
@@tomtke7351 In this case I thought it was easier to formulate just one equation. It seems more efficient to me. But sometimes I prefer the substitution method (when there are 3 unknown (X, Y and Z, for example) and also 3 given equations).
We can still solve it without factorization process. If the two numbers are x and y, we have x + y = 27 and xy = 92. Using the algebraic formulae (x - y)² = (x + b)² - 4 xy = 27² - 4 x 92 = 361 = 19². Therefore, (x - y) = 19. Given x + y = 27 and x - y = 19, our X = ½ (27+19) = 23 and y = ½ (27-19) = 4 .
Thanks for this problem. I got it right but think I may have taken too long a path. I approached it by creating two equations: x+y=27 and xy=92. Then I solved the first equation for x (x=27-y), then plugged that into the second equation: y(27-y)=92. Then I set that equal to 0 and used the quadratic formula to solve. It worked, but I worry on an exam I'd never get through all the questions if I solve this way. Among other things, I just didn't quickly see that 23 and 4 would get me what I needed so went though the whole quadratic equation. Thoughts about how to solve faster?
this is such an easy problem.. u just use estimates to solve it tha factors of 92 is 2*2*13 so the only solution is x+4=27 (2*2 factors), (the factor 13 is to big 13^2 is a 3 digit number (169), all numbers squared that start with a sequence of number greater than 3 will result in an nuber twice the number of digits (2 * number of digits) ex solve: 5000 split into two numbers that eacuals 1223436 when multiplied together... .. end num digits = 7, 5000-x can never be over 3 digits long and under 333 (its 258) if the numbers r irational or decimal...this (above) method just gives an estimate
Since both the product and the sum are natural, the numbers must be natural as well. Fractions can sum up to naturals, but they don't multiply to naturals. And roots can multiply to naturals, but they don't sum up to naturals: 1/2 + 1/2 = 1; 1/2 ⋅ 1/2 = 1/4 √3 ⋅ √3 = 3; √3 + √3 = 2 ⋅ √3
Over here spelling correctly is needed even in maths classes. If a teacher spelt separate like that especially when being inspected it would be a serious mistake and you might have the lesson classed as inadequate. In fact it fails on unclear wording, separate into two parts? how?
Well, we're obviously not looking for factors of 27 since that would give a product of 27. We can't be doing something tricky like x-y=27, that would give a negative product. That only leaves x+y. What I found ridiculous was that the original question boils down to finding factors of 92 that sum to 27, a small subset of solving the quadratic equation which it was expanded to. A bunch of extra steps for no purpose.
Your approach is overly complicated. As others have noted it's simple to find the prime factors. We can see 92 is even so we can divide by 2. This give us 46, also divisible by 2 so we get 23. 23 is prime so were done. 2*2 = 4, 4*23 =92, 4+23=27
I did it by finding the factors of 92. Much simpler. 92/2 = 46; 46/2 = 23 (prime). Factors are 2, 2, 23. For our problem solution 2x2 = 4. 4x23 = 92. 4 +23 = 27. QED
Don't ever make a problem more difficult that it has to be. Solving a quadratic to get this solution is unnecessarily complex.
We have to set up an equation. So X is 'number 1' and 27 - X is 'number 2'. Then you get: X(27 - X) = 92 => 27X - X^2 = 92 => - X^2 + 27X - 92 = 0 => X^2 - 27X + 92 = 0 (everything divided by -1) => (X - 23)(X -4) = 0 => X 1 = 23 and X 2 = 4. So the numbers are 23 and 4 (23 + 4 = 27 and 23 x 4 = 92).
I started by finding the factors of 92.... 2*2*23
Did the same. Started with:
92/2=46
46/2=23
23*4=92
23+4=27
done.
Literally took 3 seconds.
Same here
I think in this case, using a quadratic equation for this particular problem is like using a bucket-loader to dig a fence post hole. I set up a table of terms that added up to 27: 1+26 - nope; 2 +25 - nope; 3 + 24-nope; 4+23-bingo! Eight quick steps, one solution. I did assume that it would be a single-digit number, and a double-digit number.
two equations in two unknowns:
eq.1 N.1 + N.2 = 27 *
eq.2 N.1 × N.2 = 92
-----------------------------
* how else can you separate the number 27 into 2 parts unless you consider 27 to be a collection of integer digits from 0 to 27? The other alternative to dividing 27 into two parts is '2' and '7' who's product /= 92.
----------------------------
eq.1 N.1 + N.2 = 27
eq.2 N.1 × N.2 = 92
eq.1 N.1 + N.2 = 27
N.1 = 27 - N.2
eq.2 N.1 × N.2 = 92
(27 - N.2)N.2 = 92
27N.2 - N.2^2 = 92
N.2^2 - 27N.2 + 92 = 0
quad equation:
(-b +/- sqrt(b^2 - 4ac))/2a
here:
a = 1
b = -27
c = 92
(-b +/- sqrt(b^2 - 4ac))/2a
(27 +/-sqrt((-27)^2-4(1)(92) / 2(1)
(27 +/-sqrt(729-368)) / 2
(27 +/sqrt(361))/2
(27 +/-19)/2
rt.1 = (27 +19)/2
= (46)/2
= 23
rt.2 = (27-19)/2
= 8/2
= 4
sub conclude
N.2 = 23 or 4
eq.1 N.1 + N.2 = 27
for N.2 = 23
N.1 + 23 = 27
N.1 = 4
for N.2 = 4
N.1 + 4 = 27
N.1 = 23
the two parts of 27 are
4 & 23
verify:
eq.1 N.1 + N.2 = 27
eq.2 N.1 × N.2 = 92
eq.1 N.1 + N.2 = 27
4 + 23 =? 27
27 == 27 ✔️✔️✔️
eq.2 N.1 × N.2 = 92
4 × 23 =? 92
92 == 92 ✔️✔️✔️
This seems too complicated to me. Why did you not choose the equation X(27 - X) = 92? So much easier to solve and much faster too. :)
@@Kleermaker1000 great your mind can eliminate steps... in fact I'll observe that maturity in algebra is such. We ALL know to solve
x+6 = 2x +8
is
-x = 2
Without doing
"Subtract 'x' from both sides."
"Subtract '6' from both sided."
The process of being highly structured helps eliminate mistakes. School math is a combination of "how to" and "how fast." Actual employed mathematicians leans less on speed and more on process and ACCURACY.
@@tomtke7351 In this case I thought it was easier to formulate just one equation. It seems more efficient to me. But sometimes I prefer the substitution method (when there are 3 unknown (X, Y and Z, for example) and also 3 given equations).
We can still solve it without factorization process. If the two numbers are x and y, we have x + y = 27 and xy = 92. Using the algebraic formulae (x - y)² = (x + b)² - 4 xy = 27² - 4 x 92 = 361 = 19². Therefore, (x - y) = 19. Given x + y = 27 and x - y = 19, our X = ½ (27+19) = 23 and y = ½ (27-19) = 4 .
Thanks for this problem. I got it right but think I may have taken too long a path. I approached it by creating two equations: x+y=27 and xy=92. Then I solved the first equation for x (x=27-y), then plugged that into the second equation: y(27-y)=92. Then I set that equal to 0 and used the quadratic formula to solve. It worked, but I worry on an exam I'd never get through all the questions if I solve this way. Among other things, I just didn't quickly see that 23 and 4 would get me what I needed so went though the whole quadratic equation. Thoughts about how to solve faster?
this is such an easy problem..
u just use estimates to solve it tha factors of 92 is 2*2*13 so the only solution is x+4=27 (2*2 factors), (the factor 13 is to big 13^2 is a 3 digit number (169), all numbers squared that start with a sequence of number greater than 3 will result in an nuber twice the number of digits (2 * number of digits)
ex solve: 5000 split into two numbers that eacuals 1223436 when multiplied together...
.. end num digits = 7, 5000-x can never be over 3 digits long and under 333 (its 258)
if the numbers r irational or decimal...this (above) method just gives an estimate
Well, this much shorter way of doing it I factorised using prime numbers.
Sorta like mashing potatoes with a steam roller(?)
For this example anyway.
Separate, not seperate.
i do that often myself
Why use quadratics when you can use mental arithmetic? 92/2 = 46, 92/3 = 30ish, 92/4 = 23. QED.
Please correct spelling of separate or use spell check . Thank you
I factorized 92 and got 2, 2 and 23. 4+23= 27 so answer found. Admittedly this won't work if the numbers in question are not integers.
Since both the product and the sum are natural, the numbers must be natural as well. Fractions can sum up to naturals, but they don't multiply to naturals. And roots can multiply to naturals, but they don't sum up to naturals:
1/2 + 1/2 = 1; 1/2 ⋅ 1/2 = 1/4
√3 ⋅ √3 = 3; √3 + √3 = 2 ⋅ √3
I went off into the weeds because I gave up on factoring. 27±19/2 or 23,4 but it took much longer than it should have. lol
Got it, but as they say in sports it was ugly. ;)
Now let's watch how do it the right way.
With all due respect, I think you could say the same things in one-tenth of the time or even less! More people would watch your videos.
17 minutes to solve something I've just done in 1 minute using an equation? I'm not even going to see how it was done in such a long way.
Oh 23 x 4 = 92. That took me less than 30 secs or less. At 2:45 am, cant keep my eyes open.
4&23
Seperate?
(x+3x-3) (x+2x-2)
4x23 just did it in my head
I did quadratic equation and I got +4 and +23.... Didn't factor in the minus to get the negative 27🤔
((x+8x-12) (x+1x-11)
4 and 23
23 and 4
it's Beautiful ❤️
Maths is beautiful 🤩😊
23 & 4
Over here spelling correctly is needed even in maths classes. If a teacher spelt separate like that especially when being inspected it would be a serious mistake and you might have the lesson classed as inadequate. In fact it fails on unclear wording, separate into two parts? how?
Well, we're obviously not looking for factors of 27 since that would give a product of 27. We can't be doing something tricky like x-y=27, that would give a negative product. That only leaves x+y. What I found ridiculous was that the original question boils down to finding factors of 92 that sum to 27, a small subset of solving the quadratic equation which it was expanded to. A bunch of extra steps for no purpose.
23 and 4. Did this by trial and error.
Me too 15 seconds
Algo boost!
92-27=65. (65+27)=9 2
4x23
The numbers 23 and 4
The numbers are 4 and 23.
Your approach is overly complicated. As others have noted it's simple to find the prime factors. We can see 92 is even so we can divide by 2. This give us 46, also divisible by 2 so we get 23. 23 is prime so were done. 2*2 = 4, 4*23 =92, 4+23=27
23 x 4 = 92
I just use the quadratic formula. 😊
Although I appreciate your posts, they're waaaaay too wordy! You take 10 min to say what can be done in one...
I expected someone would say this...
4 and 23