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Solve 3x^2 + 18x + 15 = 0 by completing the square.

A: 3(x + 3)^2 = 6; x = −3 + √2, x = −3 − √2

B: 3(x + 9)^2 = 12; x = −11, x = −7

C: 3(x + 3)^2 = 12; x = −5, x = −1

D: 3(x + 9)^2 = 228; x = −9 + √76, x = −9 − √76



My Math:

3x^2 + 18x + 15 = 0

3(x^2 + 18x + ) + 15 = 0

3(x^2 + 18x + 81) + 15 - 81 = 0

(x^2 + 18x + 81) - 228 = 0


What do I do now? Can someone help me please?

Verified Answer :

zimmah

You made a mistake in your second line.

You put \(3x^{2} +18x\) in brackets, dividing the \(3 x^{2}\) by 3, but you didn't divide the 18x by 3.  

So you ended up with 3 times 18x in the end.    

You should divide the 18x by 3 too, so you end up with \(3( x^{2} +6x)+15=0\)

Answer:

c

Step-by-step explanation:

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