Question

Find all the zeroes of 8x 3
– 6x2
– 7x

Answer 1

I don't understand it is it x or multiple sing

Answer 2

Answer:

$x=0,\frac{3+\sqrt{65}}{8},\frac{3-\sqrt{65}}{8}$

Step-by-step explanation:

Given : Equation- $8x^3-6x^2-7x=0$

To find : The zeros of the equation

Solution :

$8x^3-6x^2-7x=0$

$x(8x^2-6x-7)=0$

$x=0,(8x^2-6x-7)=0$

One of the root is x=0

$8x^2-6x-7=0$

Solving by discriminant method,

General form - $ax^2+bx+c=0$

$D=b^2-4ac$

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

Equation is $8x^2-6x-7=0$

where a=8 , b=-6, c=-7

$D=b^2-4ac$

$D=(-6)^2-4(8)(-7)$

$D=36+224$

$D=260$

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

$x=\frac{-(-6)\pm\sqrt{260}}{2(8)}$

$x=\frac{6\pm2\sqrt{65}}{16}$

$x=\frac{3\pm\sqrt{65}}{8}$

Therefore, The other roots are $x=\frac{3+\sqrt{65}}{8},\frac{3-\sqrt{65}}{8}$