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Completing the square when coefficient >1

Say if I completed the square and got like 5(x+2)²+3

How would I solve it to find x? It's just the number in front that confuses me, would I just find the sqrt of the brackets like normal?

Thanks!

Reply 1

Super199

Original post by Qaiys

Say if I completed the square and got like 5(x+2)²+3

How would I solve it to find x? It's just the number in front that confuses me, would I just find the sqrt of the brackets like normal?

Thanks!

5(x+3)^2+3=0

5(x+3)^2=-3

(x+3)^2=\frac{-3}{5}

There are no real solutions for this but you would take the square root of both sides to get x if the RHS was positive.

Reply 2

white_o

Original post by Super199

5(x+3)^2+3=0

5(x+3)^2=-3

(x+3)^2=\frac{-3}{5}

There are no real solutions for this but you would take the square root of both sides to get x if the RHS was positive.

Okay, thanks! :smile:

Reply 3

TenOfThem

Original post by Qaiys

Say if I completed the square and got like 5(x+2)²+3

How would I solve it to find x? It's just the number in front that confuses me, would I just find the sqrt of the brackets like normal?

Thanks!

If, however, you had

5(x+2)^2 - 3 = 0

Then

5(x+2)^2 = 3

(x+2)^2 =\frac{3}{5}

x+2 = \pm\sqrt{\frac{3}{5}}

x = -2 \pm \sqrt{\frac{3}{5}}

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Completing the square when coefficient >1

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