Solve :

$\sqrt{2x + 9} + x = 13$

Given,

$\Rightarrow \sqrt{2x + 9} + x = 13 \\[1em] \Rightarrow \sqrt{2x + 9} = 13 - x \\[1em] \text{Squaring both sides, we get :} \\[1em] \Rightarrow (\sqrt{2x + 9})^2 = (13 - x)^2 \\[1em] \Rightarrow 2x + 9 = 169 - 26x + x^2 \\[1em] \Rightarrow x^2 - 26x - 2x + 169 - 9 = 0 \\[1em] \Rightarrow x^2 - 28x + 160 = 0 \\[1em] \Rightarrow x^2 - 20x - 8x + 160 = 0 \\[1em] \Rightarrow x(x - 20) - 8(x - 20) = 0 \\[1em] \Rightarrow (x - 20)(x - 8) = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = 8$

Substituting value of x = 20, in L.H.S of equation $\sqrt{2x + 9} + x = 13$, we get:

$\Rightarrow \sqrt{2(20) + 9} + 20 \\[1em] \Rightarrow \sqrt{40 + 9} + 20 \\[1em] \Rightarrow \sqrt{49} + 20 \\[1em] \Rightarrow 7 + 20 = 27$

L.H.S ≠ R.H.S

x = 20 is not valid

Substituting value of x = 8, in L.H.S of equation $\sqrt{2x + 9} + x = 13$, we get:

$\Rightarrow \sqrt{2(8) + 9} + 8 \\[1em] \Rightarrow \sqrt{16 + 9} + 8 \\[1em] \Rightarrow \sqrt{25} + 8 \\[1em] \Rightarrow 5 + 8 \\[1em] \Rightarrow 13$

L.H.S = R.H.S

x = 8 is valid

Hence, x = 8.