+447 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.
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
+26367 Find \(a\) such that \(ax^2+12x+9\) is the square of a binomial.
\(\begin{array}{|rcll|} \hline ax^2+12x+9 &=& a(x-x_0)^2 \\\\ ax^2+12x+9 &=& a(x^2-2xx_0+x_0^2) \\ ax^2+12x+9 &=& ax^2-2xx_0a+x_0^2a \\ 12x+9 &=& -2xx_0a+x_0^2a \quad | \quad \text{compare: } 12 = -2x_0a,\ 9 = x_0^2a \\ \hline && \begin{array}{|rcll|} \hline 12 &=& -2x_0a \\ 6 &=& -x_0a \\ \mathbf{x_0} &=& \mathbf{- \dfrac{6}{a} } \\\\ 9 &=& x_0^2a \\ 9 &=& \left(- \dfrac{6}{a}\right)^2a \\ 9 &=& \dfrac{36a}{a^2} \\ 9 &=& \dfrac{36 }{a } \\ a &=& \dfrac{36 }{9 } \\ \mathbf{a} &=& \mathbf{4} \\ \hline \end{array} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline x_0 &=& -\dfrac{6}{a} \quad | \quad a = 4 \\\\ x_0 &=& -\dfrac{6}{4} \\\\ \mathbf{x_0} &=& \mathbf{-\dfrac{3}{2} } \\ \hline && a(x-x_0)^2 \\ &=& 4\left(x+\dfrac{3}{2}\right)^2 \\ &=& \left(2\left(x+\dfrac{3}{2}\right)\right)^2 \\ &=& \left( 2x+2\cdot\dfrac{3}{2} \right)^2 \\ &=& ( 2x+3 )^2 \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline \mathbf{4x^2+12x+9} &=& \mathbf{( 2x+3 )^2} \\ \hline \end{array}\)
