Question

The quadratic equation abx2 + acx + b(bx + c) = 0 has non-zero equal and rational roots. The values of a and c respectively cannot be equal to (ab ÄâÄ 0) , find the value of a and c

Answer 1

The quadratic equation $Ax^{2}+Bx+C =0$ has roots,

$x=\frac{-B+\sqrt{B^{2}-4AC}}{2A}$

$x=\frac{-B-\sqrt{B^{2}-4AC}}{2A}$

Given: $abx^{2}+acx+b(bx+c)=0\\abx^{2}+acx+b^{2}x+bc=0\\abx^{2}+(ac+b^{2})x+bc=0$

Here,

A = ab ; B = ac+b² ; C =bc

Applying the values, we have

$x=\frac{-(ac+b^{2})+\sqrt{(ac+b^{2})^{2}-4ab^{2}c}} {2ab}\\\\=\frac{-ac-b^{2} + \sqrt{a^{2}c^{2}+2ab^{2}c+b^{4}-4ab^{2}c} }{2ab}\\\\=\frac{-ac-b^{2} + \sqrt{a^{2}c^{2}-2ab^{2}c+b^{4}} }{2ab}\\\\=\frac{-ac-b^{2} + (ac-b^{2}) }{2ab}\\\\=\frac{-2b^{2}}{2ab}\\\\x=\frac{-b}{a}$

$x=\frac{-(ac+b^{2})-\sqrt{(ac+b^{2})^{2}-4ab^{2}c}} {2ab}\\\\=\frac{-ac-b^{2} - \sqrt{a^{2}c^{2}+2ab^{2}c+b^{4}-4ab^{2}c} }{2ab}\\\\=\frac{-ac-b^{2} - \sqrt{a^{2}c^{2}-2ab^{2}c+b^{4}} }{2ab}\\\\=\frac{-ac-b^{2} - (ac-b^{2}) }{2ab}\\\\=\frac{-2ac}{2ab}\\\\x=\frac{-c}{b}$

∴$x=\frac{-b}{a} ; x=\frac{-c}{b}$