Question

Find the values of the following expressions:
(i) 16x² + 24x + 9, when $x=\frac{7}{4}$
(ii) 64x² + 81y² + 144xy, when x = 11 and $y=\frac{4}{3}$
(iii) 81x² + 16y² − 72xy, when $x=\frac{2}{3}$ and $y=\frac{3}{4}$

Answer 1

(i) Let us consider the following expression:

$16x^2+24x+9$

Now

$16x^2+24x+9={\left(4x+3\right)}^2$ (Using identity ${\left(a+b\right)}^2=a^2+2ab+b^2$)
$$\begin{gathered} \Rightarrow 16x^2+24x+9={\left(4\times \frac{7}{4}+3\right)}^2(\mathrm{Substituting}x=\frac{7}{4}) \\ \Rightarrow 16x^2+24x+9={\left(7+3\right)}^2 \\ \Rightarrow 16x^2+24x+9={10}^2 \\ \Rightarrow 16x^2+24x+9=100 \end{gathered}$$


(ii) Let us consider the following expression:

$64x^2+81y^2+144xy$

Now

$64x^2+81y^2+144xy={\left(8x+9y\right)}^2$ (Using identity ${\left(a+b\right)}^2=a^2+2ab+b^2$)
$$\begin{gathered} \Rightarrow 64x^2+81y^2+144xy={\left[8\left(11\right)+9\left(\frac{4}{3}\right)\right]}^2(\mathrm{Substituting}x=11\mathrm{and}y=\frac{4}{3}) \\ \Rightarrow 64x^2+81y^2+144xy={\left[88+12\right]}^2 \\ \Rightarrow 64x^2+81y^2+144xy={100}^2 \\ \Rightarrow 64x^2+81y^2+144xy=10000 \end{gathered}$$

(iii) Let us consider the following expression:

$81x^2+16y^2-72xy$

Now

$81x^2+16y^2-72xy={\left(9x-4y\right)}^2$ (Using identity ${\left(a+b\right)}^2=a^2-2ab+b^2$)
$$\begin{gathered} \Rightarrow 81x^2+16y^2-72xy={\left[9\left(\frac{2}{3}\right)-4\left(\frac{3}{4}\right)\right]}^2(\mathrm{Substituting}x=\frac{2}{3}\mathrm{and}y=\frac{3}{4}) \\ \Rightarrow 81x^2+16y^2-72xy={\left[6-3\right]}^2 \\ \Rightarrow 81x^2+16y^2-72xy=3^2 \\ \Rightarrow 81x^2+16y^2-72xy=9 \end{gathered}$$