Question

√x^2 -4 +√x^2 +5x + 6=√3x^2 +13x +14 solve

Answer 1

Given:

√x^2 -4 +√x^2 +5x + 6=√3x^2 +13x +14

To find:

√x^2 -4 +√x^2 +5x + 6=√3x^2 +13x +14 solve

Solution:

From given, we have,

√x^2 - 4 +√x^2 + 5x + 6 = √3x^2 + 13x + 14

squaring on both the sides, we have,

(√x^2 - 4 +√x^2 + 5x + 6)² = (√3x^2 + 13x + 14)²

$\left(\sqrt{x^2-4}+\sqrt{x^2+5x+6}\right)^2=\left(\sqrt{3x^2+13x+14}\right)^2\\\\2x^2+5x+2\sqrt{x^2-4}\sqrt{x^2+5x+6}+2=3x^2+13x+14\\\\2\sqrt{x^2-4}\sqrt{x^2+5x+6}+2=x^2+8x+14\\\\2\sqrt{x^2-4}\sqrt{x^2+5x+6}=x^2+8x+12$

again squaring on both the sides, we get,

$\left(2\sqrt{x^2-4}\sqrt{x^2+5x+6}\right)^2=\left(x^2+8x+12\right)^2\\\\4x^4+20x^3+8x^2-80x-96=x^4+16x^3+88x^2+192x+144$

(4x⁴ - x⁴) + (20x³ - 16x³) + (8x² - 88x²) + (-80x - 192x)² + (-96 - 144) = 0

3x⁴ + 4x³ - 80x² - 272x - 240 = 0

upon solving the above equation, we get,

x = -2, x = -10/3, x = 6

only x = -2 and x = 6 are true.

Therefore, the solution is x = -2, 6