Find the roots of the equation \(\frac{(x+3)}{(x+2)}\) = \(\frac{(3x-7

1.	Find the roots of the equation \(\frac{(x+3)}{(x+2)}\) = \(\frac{(3x-7)}{(2x-3)}\), x ≠ -2, \(\frac{3}{2}\).
Answer» Given, \(\frac{(x+3)}{(x+2)}\) = \(\frac{(3x-7)}{(2x-3)}\) On cross-multiplying we get, (x + 3)(2x – 3) = (x + 2)(3x – 7) ⇒ 2x²– 3x + 6x – 9 = 3x²– x – 14 ⇒ 2x²+ 3x – 9 = 3x²– x – 14 ⇒ x²– 3x – x – 14 + 9 = 0 ⇒ x²– 5x + x – 5 = 0 ⇒ x(x – 5) + 1(x – 5) = 0 ⇒ (x – 5)(x + l) – 0 Now, either x – 5 = 0 or x + 1 = 0 ⇒ x = 5 and x = -1 Thus, the roots of the given quadratic equation are 5 and -1 respectively.

Find the roots of the equation \(\frac{(x+3)}{(x+2)}\) = \(\frac{(3x-7)}{(2x-3)}\), x ≠ -2, \(\frac{3}{2}\).

Answer»

Given,

\(\frac{(x+3)}{(x+2)}\) = \(\frac{(3x-7)}{(2x-3)}\)

On cross-multiplying we get,

(x + 3)(2x – 3) = (x + 2)(3x – 7)

⇒ 2x²– 3x + 6x – 9 = 3x²– x – 14

⇒ 2x²+ 3x – 9 = 3x²– x – 14

⇒ x²– 3x – x – 14 + 9 = 0

⇒ x²– 5x + x – 5 = 0

⇒ x(x – 5) + 1(x – 5) = 0

⇒ (x – 5)(x + l) – 0

Now, either x – 5 = 0 or x + 1 = 0

⇒ x = 5 and x = -1

Thus, the roots of the given quadratic equation are 5 and -1 respectively.