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I am having trouble getting the answer using "completing the square" method. Please explain all the steps.

Eman Yalpsid
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ri_zen
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    If at all, it should be $;(3x+1)^2\color{red}-6;$ ... – DonAntonio May 29 '18 at 20:44
  • @rametics_ Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Jun 22 '18 at 21:03

5 Answers5

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Compare that

$$9x^2+6x-5$$

with

$$(3x+1)^2-6=9x^2+6x+1-6$$

To complete the square we need to start from

$$9x^2+6x+c^2$$

and select $c$ such that $2\cdot 3\cdot c=6 \implies c=1$ then

$$9x^2+6x+1=(3x+1)^2$$

then we add $-6$ in order to have

$$9x^2+6x-5=9x^2+6x+1-6=(3x+1)^2-6$$

user
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$$9x^2+6x-5=9\left(x^2+\frac23\right)-5=9\left(x+\frac13\right)^2-1-5=$$

$$=9\left(x+\frac13\right)^2-6=\left(3x+1\right)^2-6$$

DonAntonio
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We want $\color{red}{9x^2}+\color{green}{6x}-5$ to resemble $(a+b)^2$ as much as possible. The latter is $\color{red}{a^2}+\color{green}{2ab}+b^2$, so we let $a=3x$ in order to match $a^2$ with $9x^2$. If $a=3x$ and we want to match the next term $6x$ with $2ab$ ($=6xb$), we need $b=1$, of course. So now we have that $(a+b)^2=(3x+1)^2=9x^2+6x+1$. This differs from the desired $9x^2+6x-5$ by $6$, i.e., $$ 9x^2+6x-5=(9x^2+6x+1)-6=(3x+1)^2-6.$$

Hagen von Eitzen
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The first thing I would do is factor that leading "9" out of the first two terms: $9(x^2+ \frac{2}{3}x)- 5$. We can "complete the square" inside the parentheses by adding and subtracting $[(2/3)/2]^2= (1/3)^2= 1/9$: $9(x^2+ \frac{2}{3}x+ \frac{1}{9}- \frac{1}{9})- 5= 9(x- \frac{1}{3})^2- 1- 5= 9(x- \frac{1}{3})^2- 6$.

Now, put the "9" back in, allotting "3" to each part of the square :$(3x- 1)^2- 6$

user247327
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You need think in this $(a+b)^2 = a^2 + 2ab + b^2$. So, the expression $9x^2+6x-5$ you have on the first parcel $9x^2$ and the second is $6x$. Then, you can compute the square root, i.e., $\sqrt{9x^2} = 3x = a$. For the second, you have $2ab = 2.(3x)b=6x$, then $b=1$.

So, $a=3x$ and $b=1$, i.e., $(a+b)^2 = (3x+1)^2= (3x)^2+2(3x).1+1^2 = 9x^2+6x+1$.

Now, you think about the third parcel in $9x^2+6x-5$. For you get $-5$, you need to add $-6$ because $(3x+1)^2 -6= (9x^2+6x+1)-6 = 9x^2 + 6x - 5$.

William Bischoff
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