Solve the following equation by factorization:

$\sqrt{2x + 9}$ = (13 - x)

Given,

⇒ $\sqrt{2x + 9}$ = (13 - x)

Squaring both sides we get :

⇒ (2x + 9) = (13 - x)²

⇒ 2x + 9 = (13²) + (x)² - 2 × 13 × x

⇒ 2x + 9 = 169 + x² - 26x

⇒ x² - 26x + 169 - 2x - 9 = 0

⇒ x² - 28x + 160 = 0

⇒ x² - 20x - 8x + 160 = 0

⇒ x(x - 20) - 8(x - 20) = 0

⇒ (x - 20)(x - 8) = 0

⇒ (x - 20) = 0 or (x - 8) = 0      [Using Zero-product rule]

⇒ x = 20 or x = 8

⇒ x = 8.

Substituting x = 20 in the L.H.S. of this equation $\sqrt{2x + 9}$ = (13 - x)

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

Substituting x = 20 in the R.H.S. of this equation $\sqrt{2x + 9}$ = (13 - x)

⇒ 13 - x

⇒ 13 - 20

⇒ -7.

L.H.S ≠ R.H.S .

∴ x = 20 is not valid.

Hence, x = {8}.