Question

Solve each of the following equations and also check your results in each case:

$\left(xxv\right)$ $\left[\right(2x+3)+(x+5){]}^2+$

$\left[\right(2x+3)–(x+5){]}^2=10x^2+92$

Answer 1

Given:$\left[\right(2x+3)+(x+5){]}^2+[(2x+3)–(x+5){]}^2=$

$10x^2+92$

$[3x+8{]}^2+[x–2{]}^2=10x^2+92$

Using the formulae

$(a+b{)}^2=a^2+b^2+2ab$

$(a-b{)}^2=a^2+b^2-2ab$

$9x^2+48x+64+x^2–4x+4=10x^2+92$

$9x^2–10x^2+x^2+48x–4x=92–64–4$

$44x=24$

$x=\frac{24}{44}$

$=\frac{6}{11}$

To check: $\left[\right(2x+3)+(x+5){]}^2+[(2x+3)–(x+5){]}^2$

$=10x^2+92$ for $x=\frac{6}{11}$

$\text{L.H.S}=\left[\right(2x+3)+(x+5){]}^2+[(2x+3)–(x+5){]}^2$

$={[2\left(\frac{6}{11}\right)+3+\left(\frac{6}{11}\right)+5]}^2+{[2\left(\frac{6}{11}\right)+3–\left(\frac{6}{11}\right)–5]}^2$

$={[\left(\frac{12}{11}\right)+3+\left(\frac{6}{11}\right)+5]}^2+{[\left(\frac{12}{11}\right)+3–\left(\frac{6}{11}\right)–5]}^2$

$={[\left(\frac{12+33}{11}\right)+\left(\frac{6+55}{11}\right)]}^2+{[\left(\frac{12+33}{11}\right)–\left(\frac{6+55}{11}\right)]}^2$

$={[\left(\frac{45}{11}\right)+\left(\frac{61}{11}\right)]}^2+{[\left(\frac{45}{11}\right)–\left(\frac{61}{11}\right)]}^2$

$={\left(\frac{106}{11}\right)}^2+{(-\frac{16}{11})}^2$

$=\frac{11236}{121}+\frac{256}{121}$

$=\frac{11492}{121}$

$\text{R.H.S}=10x^2+92$

$=10{\left(\frac{6}{11}\right)}^2+92$

$=\left(\frac{360}{121}\right)+92$

$=\frac{360+11132}{121}$

$=\frac{11492}{121}$

$\therefore \text{L.H.S = R.H.S}$

Hence, the given equation is verified for $x=\frac{6}{11}.$