Answer to Question #195434 in Algebra for Ayesha Babar

Question

find x : 8x²+3x-6=0

Answer 1

$F(x)=0$

$F(x)=8x^2+3x-6=0$

Now we can not find its root by normal factorization method;so we use here Shree Dharacharya Formula.

For this compare the equation $8x^2+3x-6$ $=0$ by $ax^2+bx+c=0$

From above comparison we have

$a=8,b=3,c=-6$

Hence root $=$

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Putting the values of a,b,c in the above equation.

$x=\frac{-3\pm \sqrt{(-3)^2-4(8)(-6)}}{2\times8}$

$x=\dfrac{-3\pm\sqrt{9+192}}{16}$

$x=\dfrac{-3\pm\sqrt{201}}{16}$

$x=\dfrac{-3\pm13.85}{16}$

$x=\dfrac{-3+13.85}{16},\dfrac{-3-13.85}{16}$

$x=0.68,-1.05$

So the two roots of the equation are $0.68,-1.05$ .