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August 23rd, 2018, 12:19 AM  #1 
Newbie Joined: Aug 2018 From: US Posts: 1 Thanks: 0  Completing the square
I need help fixing this equation. The problem is x^28x+4=0 and the answer is 42\3, 4+2\3 in words is 4 minus two square root three, 4 plus two square root three, but my answer is 42\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 09:26 PM. 
August 23rd, 2018, 02:06 AM  #2 
Member Joined: Oct 2017 From: Japan Posts: 62 Thanks: 3 
I would use an easier method to solve it. 
August 23rd, 2018, 06:17 AM  #3  
Senior Member Joined: May 2016 From: USA Posts: 1,306 Thanks: 549  Quote:
$x^2  8x + 4 = 0 \implies x^2  8x = \ 4 \implies x^2 + 2 \left ( \dfrac{\ 8}{2} \right ) * x = \ 4 \implies$ $x^2 + 2(\ 4)x + (\ 4)^2 = \ 4 + (\ 4)^2 = 16  4 = 12 \implies$ $(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. Last edited by skipjack; August 23rd, 2018 at 09:25 PM.  
August 23rd, 2018, 02:38 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,684 Thanks: 658 
$x^28x+4=(x4)^2+k=x^28x+16+k$, so $k=12$. Solution is $x=4\pm \sqrt{12}$.

August 23rd, 2018, 10:08 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,285 Thanks: 1968 
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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