Question

Solve for x: abx^2 + (b^2 - ac)x - bc= 0

Answer 1

Answer:

The roots are $x=\frac{c}{b},\frac{-b}{a}$  

Step-by-step explanation:

Given : The quadratic equation - $abx^2 + (b^2 - ac)x - bc= 0$

To find : Solve for x?

Solution :  

Using quadratic formula,

General form - $ax^2+bx+c=0$ $D=b^2-4ac$  

Solution is $x=\frac{-b\pm\sqrt{D}}{2a}$  

Equation is $abx^2 + (b^2 - ac)x - bc= 0$

where, a=ab , $b=b^2-ac$ , c=-bc

$D=b^2-4ac$

$D=(b^2-ac)^2-4(ab)(-bc)$

$D=b^4+a^2c^2-2b^2ac+4b^2ac$

$D=(b^2+ac)^2$

Solution is $x=\frac{-b\pm\sqrt{D}}{2a}$

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

$x=\frac{-b^2+ac\pm(b^2+ac}{2ab}$  

$x=\frac{-b^2+ac+b^2+ac}{2ab},\frac{-b^2+ac-b^2-ac}{2ab}$  

$x=\frac{2ac}{2ab},\frac{-2b^2}{2ab}$  

$x=\frac{c}{b},\frac{-b}{a}$  

Therefore, The roots are $x=\frac{c}{b},\frac{-b}{a}$  

Answer 2

Given:

abx^2 + (b^2 - ac)x - bc = 0.

To Find:

The value of x.

Solution:

abx^2 + b^2x - acx - bc = 0.

bx (ax + b) - c (ax + b) = 0.

(bx - c)(ax + b) = 0.

x = c/b or,

x = -b/a.

Hence, the value of x is c/b or -b/a.