When we take the sqrt of 4a^2 shouldn't it be + or - 2a, not just 2a?

anytime you take the square root of an expression, it is considered the principal root - that means the value is only ever positive... when you have an equation and take the square root of both sides, that is when you can get two values... (you can get as many values as makes the equation true) ex: x^2 = 4 √(x^2) = √4 |x| = 2 x = ±2

In step 8 the square root on the right hand side is +/-. Why is the square root on the left hand side not also +/-?

on the left hand side x + (b/2a) is squared so the root cancels out

How can we multiply by 4a^2 in step 6, without affecting the left side of the equation?

What they did in step 6 was multiply - c/a by 4a/4a. The reason you can do this is because 4a/4a is the same thing as 1, so multiplying by it doesn't change any values. In other words, -c/a has the same value as -4ac/(4a^2)

n step 8 the square root on the right hand side is +/-. Why is the square root on the left hand side not also +/-?

That is a great question! I thought I knew algebra, but I never noticed that and it took me a little minute to work out! Only one side of the equation needs a +/- sign because if you multiply the equation by -1 you can get to any combination of negatives and positives while only putting the +/- sign on one side. I know that sounds confusing, so let's simplify things. Say, for example, we're using the equation a = ±b. The two possible situations are _a = b_ and _a = -b_, but multiplying those by -1 gives you _-a = -b_ and _-a = b_, meaning that either side could be positive or negative in any combination with the +/- sign on only one side. Terrific question! I hope this helps, and remember that you can learn anything!

can someone help me with 4x^2+11x-20=0 I solved everything expect I got stuck on the square root of 441

You need to break 441 down into prime factors to simplify the square root. Learn the divisibility tests -- see the video at this link: https://www.khanacademy.org/math/pre-algebra/pre-algebra-factors-multiples/pre-algebra-divisibility-tests/v/divisibility-tests-for-2-3-4-5-6-9-10 If you know these test, you would recognize that 441 is divisible by 9. 441 / 9 = 49 So, prime factors of 441 = 3 * 3 * 7 * 7 And sqrt(441) = 21 Hope this helps.

I require some help with understanding how -b/2a derives the x-coordinate of the vertex of a parabola. Thanks in advance!

I know of two ways to understanding it. First, using the vertex formula: y = a(x – h)^2 + k, where "h" is the vertex. Put the general equation y = ax^2 + bx + c into the vertex form and you will find that "h" will equal -b/2a. I'll leave the work up to you. Second, since quadratics in the general form (y = ax^2 + bx + c) are symmetric over a vertical line through the vertex, we can use the two roots of the quadratic formula and average them to find the x-coordinate of the vertex (visualize a quadratic graph and you will see why this is true). So if you find the average of the two roots: [-b + sqrt(b^2-4ac)]/2a and [-b - sqrt(b^2-4ac)]/2a // it will be -b/2a. (I again, will leave the work up to you.)

why don't we write +/- when we take the square root of 4a^2 ?

Look at answer to similar question 8 days ago from emilio12medina. The idea is that you do not need two sets of ± signs, they end up cancelling each other out.

on step 4, why adding (b^2/4a^2) to both sides?

To complete the square, you divide the coefficient of the x term by 2 (b/2a) and square this to get b^2/4a^2. So you need this term to complete the square. If you do it to the left side in order to complete the square, you either have to subtract it on the left or add it to the right side of the equation to keep it balanced.

Main content

Algebra 1

Course: Algebra 1 > Unit 14

Lesson 8: More on completing the square

Quadratic formula proof review

Google Classroom

A text-based proof (not video) of the quadratic formula

The quadratic formula says that

x, equals, start fraction, minus, start color #e07d10, b, end color #e07d10, plus minus, square root of, start color #e07d10, b, end color #e07d10, squared, minus, 4, start color #7854ab, a, end color #7854ab, start color #e84d39, c, end color #e84d39, end square root, divided by, 2, start color #7854ab, a, end color #7854ab, end fraction

for any quadratic equation like:

start color #7854ab, a, end color #7854ab, x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #e84d39, c, end color #e84d39, equals, 0

If you've never seen this formula proven before, you might like to watch a video proof, but if you're just reviewing or prefer a text-based proof, here it is:

The proof

We'll start with the general form of the equation and do a whole bunch of algebra to solve for x. At the heart of the proof is the technique called start color #11accd, start text, c, o, m, p, l, e, t, i, n, g, space, t, h, e, space, s, q, u, a, r, e, end text, end color #11accd. If you're unfamiliar with this technique, you may want to brush up by watching a video.

Part 1: Completing the square

\begin{aligned} \purpleD{a}x^2 + \goldD{b}x + \redD{c} &= 0&(1)\\\\ ax^2+bx&=-c&(2)\\\\ x^2+\dfrac{b}{a}x&=-\dfrac{c}{a}&(3)\\\\ \blueD{x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}}&\blueD{=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}&(4)\\\\ \blueD{\left (x+\dfrac{b}{2a}\right )^2}&\blueD{=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}&(5) \end{aligned}

Part 2: Algebra! Algebra! Algebra!

Remember, our goal is to solve for x.

\begin{aligned} \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2}{4a^2}-\dfrac{c}{a}&(5) \\\\ \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2}{4a^2}-\dfrac{4ac}{4a^2} &(6)\\\\ \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2-4ac}{4a^2}&(7)\\\\ x+\dfrac{b}{2a}&=\pm \dfrac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}&(8)\\\\ x+\dfrac{b}{2a}&=\pm \dfrac{\sqrt{b^2-4ac}}{2a}&(9)\\\\ x&=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}&(10)\\\\ x&=\dfrac{-\goldD{b}\pm\sqrt{\goldD{b}^2-4\purpleD{a}\redD{c}}}{2\purpleD{a}}&(11) \end{aligned}

And we're done!

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jd1311993
Posted 7 years ago. Direct link to jd1311993's post “When we take the sqrt of ...”
When we take the sqrt of 4a^2 shouldn't it be + or - 2a, not just 2a?
Button navigates to signup pageComment on jd1311993's post “When we take the sqrt of ...”
(17 votes)
Answer
- Ms Demaray
  Posted 7 years ago. Direct link to Ms Demaray's post “anytime you take the squa...”
  anytime you take the square root of an expression, it is considered the principal root - that means the value is only ever positive... when you have an equation and take the square root of both sides, that is when you can get two values... (you can get as many values as makes the equation true) ex:
  x^2 = 4
  √(x^2) = √4
  |x| = 2
  x = ±2
  Button navigates to signup page
  (6 votes)
Derrick Logan
Posted 6 years ago. Direct link to Derrick Logan's post “In step 8 the square root...”
In step 8 the square root on the right hand side is +/-. Why is the square root on the left hand side not also +/-?
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(9 votes)
Answer
- carleboyy
  Posted 6 years ago. Direct link to carleboyy's post “on the left hand side x +...”
  on the left hand side x + (b/2a) is squared so the root cancels out
  Button navigates to signup page
  (5 votes)
PythagorasLessFortunateBrother
Posted 3 years ago. Direct link to PythagorasLessFortunateBrother's post “How can we multiply by 4a...”
How can we multiply by 4a^2 in step 6, without affecting the left side of the equation?
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(5 votes)
Answer
- Bradley Reynolds
  Posted 3 years ago. Direct link to Bradley Reynolds's post “What they did in step 6 w...”
  What they did in step 6 was multiply - c/a by 4a/4a. The reason you can do this is because 4a/4a is the same thing as 1, so multiplying by it doesn't change any values.
  In other words, -c/a has the same value as -4ac/(4a^2)
  Button navigates to signup page
  (9 votes)
Jesus fan
Posted 4 months ago. Direct link to Jesus fan's post “number one victory royale...”
number one victory royale...
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(8 votes)
Answer
Raghav
Posted 6 years ago. Direct link to Raghav's post “I require some help with ...”
I require some help with understanding how -b/2a derives the x-coordinate of the vertex of a parabola. Thanks in advance!
Button navigates to signup pageComment on Raghav's post “I require some help with ...”
(1 vote)
Answer
- bsd
  Posted 6 years ago. Direct link to bsd's post “I know of two ways to und...”
  I know of two ways to understanding it.
  
  First, using the vertex formula: y = a(x – h)^2 + k, where "h" is the vertex.
  Put the general equation y = ax^2 + bx + c into the vertex form and you will find that "h" will equal -b/2a. I'll leave the work up to you.
  
  Second, since quadratics in the general form (y = ax^2 + bx + c) are symmetric over a vertical line through the vertex, we can use the two roots of the quadratic formula and average them to find the x-coordinate of the vertex (visualize a quadratic graph and you will see why this is true).
  
  So if you find the average of the two roots:
  
  [-b + sqrt(b^2-4ac)]/2a and [-b - sqrt(b^2-4ac)]/2a // it will be -b/2a. (I again, will leave the work up to you.)
  Button navigates to signup page
  (12 votes)
Santiago Ramirez
Posted 3 years ago. Direct link to Santiago Ramirez's post “n step 8 the square root ...”
n step 8 the square root on the right hand side is +/-. Why is the square root on the left hand side not also +/-?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Neil Gabrielson
  Posted 3 years ago. Direct link to Neil Gabrielson's post “That is a great question!...”
  That is a great question! I thought I knew algebra, but I never noticed that and it took me a little minute to work out!
  Only one side of the equation needs a +/- sign because if you multiply the equation by -1 you can get to any combination of negatives and positives while only putting the +/- sign on one side. I know that sounds confusing, so let's simplify things. Say, for example, we're using the equation a = ±b. The two possible situations are a = b and a = -b, but multiplying those by -1 gives you -a = -b and -a = b, meaning that either side could be positive or negative in any combination with the +/- sign on only one side. Terrific question! I hope this helps, and remember that you can learn anything!
  Button navigates to signup page
  (4 votes)
ranoosh
Posted 6 years ago. Direct link to ranoosh's post “can someone help me with ...”
can someone help me with 4x^2+11x-20=0 I solved everything expect I got stuck on the square root of 441
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Kim Seidel
  Posted 6 years ago. Direct link to Kim Seidel's post “You need to break 441 dow...”
  You need to break 441 down into prime factors to simplify the square root.
  Learn the divisibility tests -- see the video at this link: https://www.khanacademy.org/math/pre-algebra/pre-algebra-factors-multiples/pre-algebra-divisibility-tests/v/divisibility-tests-for-2-3-4-5-6-9-10
  
  If you know these test, you would recognize that 441 is divisible by 9.
  441 / 9 = 49
  So, prime factors of 441 = 3 * 3 * 7 * 7
  And sqrt(441) = 21
  Hope this helps.
  Comment on Kim Seidel's post “You need to break 441 dow...”
  (4 votes)
Qasim Hashmi
Posted 6 months ago. Direct link to Qasim Hashmi's post “why don't we write +/- wh...”
why don't we write +/- when we take the square root of 4a^2 ?
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(1 vote)
Answer
- David Severin
  Posted 6 months ago. Direct link to David Severin's post “Look at answer to similar...”
  Look at answer to similar question 8 days ago from emilio12medina. The idea is that you do not need two sets of ± signs, they end up cancelling each other out.
  Button navigates to signup page
  (4 votes)
Lyn Kang
Posted 2 years ago. Direct link to Lyn Kang's post “I tried the proof myself ...”
I tried the proof myself in a slightly different way and it didn't quite work out.

(1) ax^2 + bx + c = 0
(2) x^2 + (b/a)x + c/a = 0
(3) x^2 + (b/a)x + (b^2/4a^2) + c/a - (b^2/4a^2) = 0
(4) (x+(b/2a))^2 + 4ac/4a^2 - b^2/4a^2 = 0
(5) (x+(b/2a))^2 + (4ac-b^2)/(4a^2) = 0
(6) (x+(b/2a))^2 = -(4ac-b^2)/(4a^2)
(7) (x+(b/2a)) = -(sqrt(4ac-b^2))/2a
(8) x = -(b/2a)-(sqrt(4ac-b^2))/2a
(9) x = -(-b+-squr(4ac-b^2))/2a

My discriminant is 4ac-b^2 while the one we use is b^2-4ac. Also, I have a negative sign in front of the whole fraction which is from the second equation in step 8. Could somebody tell me where I made a mistake?
Button navigates to signup pageComment on Lyn Kang's post “I tried the proof myself ...”
(3 votes)
Answer
- David Severin
  Posted 2 years ago. Direct link to David Severin's post “At step 6 and 7 when you ...”
  At step 6 and 7 when you took the square root, the negative should have stayed inside rather than outside (taking square root would yield ± on the outside). So when you distribute the -1(4ac-b^2) you end up with b^2-4ac.
  Button navigates to signup page
  (1 vote)
Hann
Posted 5 years ago. Direct link to Hann's post “on step 4, why adding (b^...”
on step 4, why adding (b^2/4a^2) to both sides?
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(1 vote)
Answer
- David Severin
  Posted 5 years ago. Direct link to David Severin's post “To complete the square, y...”
  To complete the square, you divide the coefficient of the x term by 2 (b/2a) and square this to get b^2/4a^2. So you need this term to complete the square. If you do it to the left side in order to complete the square, you either have to subtract it on the left or add it to the right side of the equation to keep it balanced.
  Button navigates to signup page
  (5 votes)