Solve each of the following equation and also check your result in each case:

Solution:

$[(2 \mathrm{x}+3)+(\mathrm{x}+5)]^{2}+[(2 \mathrm{x}+3)-(\mathrm{x}+5)]^{2}=10 \mathrm{x}^{2}+92$

or $(3 \mathrm{x}+8)^{2}+(\mathrm{x}-2)^{2}=10 \mathrm{x}^{2}+92$

or $9 \mathrm{x}^{2}+48 \mathrm{x}+64+\mathrm{x}^{2}-4 \mathrm{x}+4=10 \mathrm{x}^{2}+92 \quad\left[(\mathrm{a}+\mathrm{b})^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab} \quad \mathrm{and}(\mathrm{a}-\mathrm{b})^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{ab}\right]$

or $10 x^{2}-10 x^{2}+44 x=92-68$

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

or $\mathrm{x}=\frac{6}{11}$

Thus, $\mathrm{x}=\frac{6}{11}$ is the solution of the given equation.

Check:

Substituting $\mathrm{x}=\frac{6}{11}$ in the given equation, we get:

L.H.S. $=\left[\left(2 \times \frac{6}{11}+3\right)+\left(\frac{6}{11}+5\right)\right]^{2}+\left[\left(2 \times \frac{6}{11}+3\right)-\left(\frac{6}{11}+5\right)\right]^{2}$

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

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

$=\frac{11492}{121}$

R.H.S. $=10 \times\left(\frac{6}{11}\right)^{2}+92=\frac{360}{121}+92=\frac{11492}{121}$

$\therefore \mathrm{L} . \mathrm{H} . \mathrm{S} .=\mathrm{R} . \mathrm{H} . \mathrm{S} .$ for $\mathrm{x}=\frac{6}{11}$