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# algebra II help

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Find $$a$$ such that $$ax^2+12x+9$$ is the square of a binomial.

Oct 20, 2019

#1
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Step 1: Square the first term of the binomial. Step 2: Multiply the first term and last term of the binomial together and then double that quantity (in other words multiply by 2). Step 3: Square the last term of the binomial.

Oct 20, 2019
#2
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the square of a binomial: (xy+b)(xy+b)=x^2y^2+2xyb+b^2

9=3^2=b

12x=2xyb

12=yb

12=3y

y=4

a=4

Oct 21, 2019
#3
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Find $$a$$  such that $$ax^2+12x+9$$ is the square of a binomial.

$$\begin{array}{|rcll|} \hline ax^2+12x+9 &=& a(x-x_0)^2 \\\\ ax^2+12x+9 &=& a(x^2-2xx_0+x_0^2) \\ ax^2+12x+9 &=& ax^2-2xx_0a+x_0^2a \\ 12x+9 &=& -2xx_0a+x_0^2a \quad | \quad \text{compare: } 12 = -2x_0a,\ 9 = x_0^2a \\ \hline && \begin{array}{|rcll|} \hline 12 &=& -2x_0a \\ 6 &=& -x_0a \\ \mathbf{x_0} &=& \mathbf{- \dfrac{6}{a} } \\\\ 9 &=& x_0^2a \\ 9 &=& \left(- \dfrac{6}{a}\right)^2a \\ 9 &=& \dfrac{36a}{a^2} \\ 9 &=& \dfrac{36 }{a } \\ a &=& \dfrac{36 }{9 } \\ \mathbf{a} &=& \mathbf{4} \\ \hline \end{array} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline x_0 &=& -\dfrac{6}{a} \quad | \quad a = 4 \\\\ x_0 &=& -\dfrac{6}{4} \\\\ \mathbf{x_0} &=& \mathbf{-\dfrac{3}{2} } \\ \hline && a(x-x_0)^2 \\ &=& 4\left(x+\dfrac{3}{2}\right)^2 \\ &=& \left(2\left(x+\dfrac{3}{2}\right)\right)^2 \\ &=& \left( 2x+2\cdot\dfrac{3}{2} \right)^2 \\ &=& ( 2x+3 )^2 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{4x^2+12x+9} &=& \mathbf{( 2x+3 )^2} \\ \hline \end{array}$$

Oct 21, 2019
edited by heureka  Oct 21, 2019