Question

Solve using "Quadratic formula"​

Answer 1

$\sf \implies \: \dfrac{x + 1}{2x + 5} = \dfrac{x + 3}{3x + 4} \\ \\ \sf \to \:( x + 1)(3x + 4) = (x + 3)(2x + 5) \\ \\ \sf \to \: 3x {}^{2} + 4x + 3x + 4 = 2x {}^{2} + 5x + 6x + 15 \\ \\ \sf \to \: x {}^{2} + 7x - 11x = 15 - 4 \\ \\ \sf \to \: x {}^{2} - 4x = 11 \\ \\ \to\boxed{x {}^{2} - 4x - 11 = 0}$

➸ By comparing this equation by ax² + bx + c = 0,

• a = 1
• b = -4
• c = -11

$\sf \: x = \dfrac{ - b + \sqrt{b {}^{2} - 4ac } }{2a} \\ \\ x = \dfrac{ - ( - 4) + \sqrt{( - 4) {}^{2} - 4(1)( - 11) } }{2(1)} \\ \\ \implies \blue{\boxed {x = \dfrac{4+ 2\sqrt{15}}{2}}}$

⇝Similarly,

$\implies \blue{\boxed {x = \dfrac{4- 2\sqrt{15}}{2}}}$

⇝Simplify the values of x, we get

x = 2 + √15 and 2 - √15