Question

The quadratic equation 8x²+12x-14 has two real roots. What is the sum of the squares of these roots?

Answer 1

The real roots are

`x=((-3+√(37)))/(4)` and
`x=((-3-√(37)))/(4)`

The sum of the squares of these roots is
`(-3)/(2)`

Explanation:

The given quadratic equation is
`8x^2+12x-14` has two real roots.

To find the roots .

`8x^2+12x-14=0`

Dividing the above equation by 2

`(1)/(2)(8x^2+12x-14)=(0)/(2)`

`4x^2+6x-7=0`

For quadratic equation
`ax^2+bx+c=0` the solution is
`x=(-b\pm√(b^2-4ac))/(2a)`

Where a and b are coefficents of
`x^2` and x respectively, c is a constant.

For given quadratic equation

a=4, b=6, c=-7

`x=(-6\pm√(6^2-4(4)(-7)))/(2(4))`

`=(-6\pm√(36+112))/(8)`

`=(-6\pm√(148))/(8)`

`=(-6\pm√(37* 4))/(8)`

`=(-6\pm√(37)*√(4))/(8)`

`=(-6\pm√(37)* 2)/(8)`

`=2((-3\pm√(37)))/(8)`

`=(-3\pm√(37))/(4)`

`x=((-3\pm√(37)))/(4)`

The real roots are

`x=((-3+√(37)))/(4)` and
`x=((-3-√(37)))/(4)`

Now to find the sum of the squares of these roots

`\left[(-3+√(37))/(4)+((-3-√(37)))/(4)\right]^2=(-3+√(37)-3-√(37))/(4)`

`=(-6)/(4)`

`=(-3)/(2)`

`\left[(-3+√(37))/(4)+((-3-√(37)))/(4)\right]^2=(-3)/(2)`

Therefore the sum of the squares of these roots is
`(-3)/(2)`