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August 22nd, 2018, 11:19 PM   #1
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Completing the square

I need help fixing this equation. The problem is x^2-8x+4=0 and the answer is 4-2\|3, 4+2\|3 in words is 4 minus two square root three, 4 plus two square root three, but my answer is 4-2\|5,4+2\|5 ..why do I have a square root of five but not square root of three? Below is my work.

Last edited by skipjack; August 23rd, 2018 at 08:26 PM.
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August 23rd, 2018, 01:06 AM   #2
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I would use an easier method to solve it.
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August 23rd, 2018, 05:17 AM   #3
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Quote:
Originally Posted by Gold Boy View Post
I need help fixing this equation. The problem is x^2-8x+4=0 and the answer is 4-2\|3, 4+2\|3 in words is 4 minus two square root three, 4 plus two square root three, but my answer is 4-2\|5,4+2\|5 ..why do I have a square root of five but not square root of three? Below is my work.
I am not sure why you are completing the square, but here goes

$x^2 - 8x + 4 = 0 \implies x^2 - 8x = -\ 4 \implies x^2 + 2 \left ( \dfrac{-\ 8}{2} \right ) * x = -\ 4 \implies$

$x^2 + 2(-\ 4)x + (-\ 4)^2 = -\ 4 + (-\ 4)^2 = 16 - 4 = 12 \implies$

$(x - 4)^2 = 12 \implies x - 4 = \pm \sqrt{12} = \pm \sqrt{4 * 3} = \pm 2 \sqrt{3} \implies $

$x = 4 \pm 2\sqrt{3}.$

Let's check.

$x = 4 \pm 2\sqrt{3} \implies x^2 - 8x + 4 = (16 \pm 2 * 4 * 2\sqrt{3}+ 4 * 3) - 8(4 \pm 2 \sqrt{3}) + 4 =$

$x = 16 + 12 + 4 - 32 \pm 16\sqrt{3} \mp 16\sqrt{3} = 32 - 32 = 0.$

EDIT: You must be very careful about signs when completing the square and when using the quadratic formula.

EDIT 2: You neglected to add 16 to BOTH sides of the equation, and then equated

$\sqrt{-\ 20}$ to $\sqrt{20}.$

So you made two basic errors.
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Last edited by skipjack; August 23rd, 2018 at 08:25 PM.
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August 23rd, 2018, 01:38 PM   #4
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$x^2-8x+4=(x-4)^2+k=x^2-8x+16+k$, so $k=-12$. Solution is $x=4\pm \sqrt{12}$.
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August 23rd, 2018, 09:08 PM   #5
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If possible, verify each step as you do it, so that any mistake is found immediately. Verifying a step isn't the same as checking that you did what you intended to, as your intention might have been incorrect.
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