Hi I'm Josh . I hope you're doing well! I solved the quadratic equation 2x^2 - 4x - 5 = 0 and wanted to bring a small issue to your attention. In the video, the solution provided was x = 1 ± (1/2)√14. However, other correct solutions can be x = 1 ± (√14)/2 or equivalently x = 1 ± √(7/2). I botg solved this using the quadratic formula and and using completing the square, I get equivalent answers. Using the Quadratic Formula: For the quadratic equation 2x^2 - 4x - 5 = 0: a = 2, b = -4, c = -5 Apply the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a) Substitute a, b, and c: x = (-(-4) ± √((-4)^2 - 4 * 2 * (-5))) / (2 * 2) Simplify: x = (4 ± √(16 + 40)) / 4 x = (4 ± √56) / 4 x = (4 ± √(4 * 14)) / 4 x = (4 ± 2√14) / 4 x = 1 ± (√14) / 2 Using Completing the Square: Start with 2x^2 - 4x - 5 = 0, divide by 2: x^2 - 2x - 5/2 = 0 Move the constant term: x^2 - 2x = 5/2 Complete the square: x^2 - 2x + 1 = 5/2 + 1 (x - 1)^2 = 7/2 Take the square root: x - 1 = ± √(7/2) x = 1 ± (√14) / 2 Both methods confirm that the correct solutions are x = 1 ± (√14) / 2, which simplifies to x = 1 ± √(7/2). I just wanted to clarfiy this, if I were wrong I'll just accept it. Thanks for understanding and you are one of the best math tutor you made math easy for me!
@@joshuagallaza_orig (sqrt14)/2 does not simplify to (sqrt7)/2. You cannot divide what is inside a square root by a constant expression. Additionally, his solution of (1/2)*sqrt14 is equivalent to (sqrt14)/2, so he is most definitely correct.
I did not say that Sir Brian was wrong; all three solutions are correct, as they are equivalent. I just wanted to clarify that. Honestly, at first, I thought Sir Brian was wrong.
Thank you for the detailed tutorial. I just encountered your page, and you gained a new subscriber. Indeed, a good understanding of factoring is important to help one gain a solid foundation in Algebra, since factoring is a valuable skill to have when dealing with quadratic functions and higher degree polynomials. I thought that there were three things that you should have also further explained. A. In; x^2 - 4 = 0. Apart from transposing 4 to the right side and taking the square root of both sides, the left side can also be factored as a difference of two squares. Thus, (x - 2)(x + 2) = 0 The two linear factors can then be solved to give the two roots of x = 2 and x = - 2. B. In showing the completing the square method, you should have gone step-by-step. You simply used your advanced knowledge to state c = (b/2)^2. People who are new to the concept might not know how this came about. They should know that the 'c' term has been transposed to the right side of the equation, and one-half the b term squared has to be added to both sides of the equation to complete the square, or to have a perfect square on the left, which can then be solved for x. C. For the graphical method, it is possible that you wanted to be brief, but I think that you should have mentioned that f(y) = 0, so that your students know that the two values of x would be the x intercepts of the graph when y = 0. Perhaps these details were not properly conveyed because of your fast-paced method. Again, for the graphical method, I think students should also be taught how to come up with a table of values and actually take a graphing sheet, and plot a quadratic function, and locate the x intercepts that are the solutions to the quadratic equation. I am sure that is how you were taught, and such an important skill should be passed on to the newer generation of students who are always very quick to plug in values in some Internet Mathematics Web site to generate a finished graph or to use a graphing calculator. Once again, thank you for sharing your invaluable knowledge. Cheers.
You think this man is great. You don't have to go to India to see a genius. If you come to south Africa you will be shocked with the mathematical talents the indians have. I am am simple Tom, Dick and Harry, an Indian from south Africa I will just use the quadratic formula. Period. This advice is free of charge 😂
One method I like you could call the sum/product method. If you have a quadratic x^2+bx+c=0, the sum of the roots is -b and the product is c. So think of the roots as r+s and r-s. The sum of the roots is 2r, so r is just-b/2. The product of the roots is r^2-s^2, so s is just the square root of r^2-c. Take the x^2-4x+1=0 example. r=4/2=2 and s is the square root of 2^2-1=3. It can be done quite easily in your head if b is even, even if the roots are irrational or complex.
You really need to add a context to this talk explaining WHY you want to solve a quadratic. Why is it important? Other than trivial reasons like "it's there" or "it will appear on a test". Instead describe where a quadratic fits within the math realm.
@@normanklein3155 nah - there's enough videos that treat math like a recipe blog. The reason this channel is so valuable is that is explicitly focused on how to do things imo
imagine that he was writing on simple glass or perspex or s.l.t. , is filmed through this glass and the video has been mirrored afterward . because his writing is obviously right handed - in direction from thumb to pinky , not reverse .
I get the perfect square example (factoring) but why there is only one solution in this case? I thought you always need 2 solutions for a quadratic equation. Can you explain that?
I'm calling foul on your first example, Mr. Mc Logan. Four only has one square root. sqrt[4] = 2. sqrt[x^2] = |x|. It follows that sqrt[x^2] = sqrt[4] implies |x| = 2. Therefore, x = +/- 2 ◼ No worries. Every Algebra teacher on the planet makes that mistake. Some of them never get over it. (If it helps, think of it as the square root is the side length of a square having area A. A negative length is absolutely nonsensical.)
Just saw this chanel. Love it
Thank you man, you made me smarter today and as I keep watching. You’ll make me smarter everyday‼️
Hi I'm Josh . I hope you're doing well! I solved the quadratic equation 2x^2 - 4x - 5 = 0 and wanted to bring a small issue to your attention.
In the video, the solution provided was x = 1 ± (1/2)√14. However, other correct solutions can be x = 1 ± (√14)/2 or equivalently x = 1 ± √(7/2). I botg solved this using the quadratic formula and and using completing the square, I get equivalent answers.
Using the Quadratic Formula:
For the quadratic equation 2x^2 - 4x - 5 = 0:
a = 2, b = -4, c = -5
Apply the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Substitute a, b, and c:
x = (-(-4) ± √((-4)^2 - 4 * 2 * (-5))) / (2 * 2)
Simplify:
x = (4 ± √(16 + 40)) / 4
x = (4 ± √56) / 4
x = (4 ± √(4 * 14)) / 4
x = (4 ± 2√14) / 4
x = 1 ± (√14) / 2
Using Completing the Square:
Start with 2x^2 - 4x - 5 = 0, divide by 2:
x^2 - 2x - 5/2 = 0
Move the constant term:
x^2 - 2x = 5/2
Complete the square:
x^2 - 2x + 1 = 5/2 + 1
(x - 1)^2 = 7/2
Take the square root:
x - 1 = ± √(7/2)
x = 1 ± (√14) / 2
Both methods confirm that the correct solutions are x = 1 ± (√14) / 2, which simplifies to x = 1 ± √(7/2). I just wanted to clarfiy this, if I were wrong I'll just accept it. Thanks for understanding and you are one of the best math tutor you made math easy for me!
@@joshuagallaza_orig (sqrt14)/2 does not simplify to (sqrt7)/2. You cannot divide what is inside a square root by a constant expression. Additionally, his solution of (1/2)*sqrt14 is equivalent to (sqrt14)/2, so he is most definitely correct.
@@KitKat-xt4ti what i meant dude is the square root of 7/2 is equal to square root of four over 2 not (sqrt14)/2 = (sqrt7)/2
@@KitKat-xt4ti u see if i divide square root of two to the sqrt of 7/2 to both the denominator and numerator, it results to sqrt of 14 over 2
I did not say that Sir Brian was wrong; all three solutions are correct, as they are equivalent. I just wanted to clarify that. Honestly, at first, I thought Sir Brian was wrong.
I love your teaching
Thank you for the detailed tutorial. I just encountered your page, and you gained a new subscriber. Indeed, a good understanding of factoring is important to help one gain a solid foundation in Algebra, since factoring is a valuable skill to have when dealing with quadratic functions and higher degree polynomials.
I thought that there were three things that you should have also further explained.
A.
In;
x^2 - 4 = 0.
Apart from transposing 4 to the right side and taking the square root of both sides, the left side can also be factored as a difference of two squares. Thus,
(x - 2)(x + 2) = 0
The two linear factors can then be solved to give the two roots of x = 2 and x = - 2.
B.
In showing the completing the square method, you should have gone step-by-step. You simply used your advanced knowledge to state c = (b/2)^2. People who are new to the concept might not know how this came about. They should know that the 'c' term has been transposed to the right side of the equation, and one-half the b term squared has to be added to both sides of the equation to complete the square, or to have a perfect square on the left, which can then be solved for x.
C.
For the graphical method, it is possible that you wanted to be brief, but I think that you should have mentioned that f(y) = 0, so that your students know that the two values of x would be the x intercepts of the graph when y = 0. Perhaps these details were not properly conveyed because of your fast-paced method.
Again, for the graphical method, I think students should also be taught how to come up with a table of values and actually take a graphing sheet, and plot a quadratic function, and locate the x intercepts that are the solutions to the quadratic equation. I am sure that is how you were taught, and such an important skill should be passed on to the newer generation of students who are always very quick to plug in values in some Internet Mathematics Web site to generate a finished graph or to use a graphing calculator.
Once again, thank you for sharing your invaluable knowledge. Cheers.
Your clear and concise ways to explain remind me of my College teacher in Cuba.
Thank you sooooo much for your videos
🇮🇶I am following you from Iraq, thanks for the clarification 😢❤
@@Oo-sd3ke same sis
@@Oo-sd3ke I'm from Nepal 🫰
how you write like that is fascinating
You’re either being a cynic or you just didn’t realize the video image was reversed. Look at the t-shirt logo.
You are good teacher
Thanks❤
Love from 🇮🇳
Sir what you think about iota
Hes truely the Best maths teacher!!!.
Mr Mohammed you haven't come across better teachers eg. In India. The street sweepers know more maths than this a.h.
You think this man is great. You don't have to go to India to see a genius. If you come to south Africa you will be shocked with the mathematical talents the indians have. I am am simple Tom, Dick and Harry, an Indian from south Africa I will just use the quadratic formula. Period. This advice is free of charge 😂
@@harrymatabal8448 yoo don't talk about your fathers here😮💨
One method I like you could call the sum/product method. If you have a quadratic x^2+bx+c=0, the sum of the roots is -b and the product is c. So think of the roots as r+s and r-s. The sum of the roots is 2r, so r is just-b/2. The product of the roots is r^2-s^2, so s is just the square root of r^2-c. Take the x^2-4x+1=0 example. r=4/2=2 and s is the square root of 2^2-1=3. It can be done quite easily in your head if b is even, even if the roots are irrational or complex.
Perfect
Completing the square can be used to derive the quadratic formula.
You really need to add a context to this talk explaining WHY you want to solve a quadratic. Why is it important? Other than trivial reasons like "it's there" or "it will appear on a test". Instead describe where a quadratic fits within the math realm.
@@normanklein3155 nah - there's enough videos that treat math like a recipe blog. The reason this channel is so valuable is that is explicitly focused on how to do things imo
Can you please let me know the name of the software that allows you to write on the screen that you are using to make this video? Thank You
@@domanicmarcus2176 isn't that an actual marker on a physical whiteboard
imagine that he was writing on simple glass or perspex or s.l.t. , is filmed through this glass and the video has been mirrored afterward .
because his writing is obviously right handed - in direction from thumb to pinky , not reverse .
Can we apply the formula of (a-b)² in the second question? How to identify
#6 Viete’s relations
I get the perfect square example (factoring) but why there is only one solution in this case? I thought you always need 2 solutions for a quadratic equation.
Can you explain that?
There are more than 1 way ?
Could you not have factorized x^2 -4
My man
How could you write like this from right to left
I'm calling foul on your first example, Mr. Mc Logan. Four only has one square root. sqrt[4] = 2.
sqrt[x^2] = |x|.
It follows that sqrt[x^2] = sqrt[4] implies |x| = 2. Therefore, x = +/- 2 ◼
No worries. Every Algebra teacher on the planet makes that mistake. Some of them never get over it.
(If it helps, think of it as the square root is the side length of a square having area A. A negative length is absolutely nonsensical.)