Question

How do you find the roots, real and imaginary, of `y=-(x-1)^2 - x^2 - 3x + 2` using the quadratic formula?

Answer 1

Required Solutions are:

`color(blue)(x = 1/2 or x = 0.5)`

`color(blue)(x = -1)`

Explanation:

We are given

`color(red)(y = - (x-1)^2 - x^2 - 3x+2)`

We need to use the quadratic formula to find the roots, real or imaginary.

The algebraic expression on the Right-hand-Side(RHS) can be simplified to ...

`color(blue)(-(x^2 - 2x +1) - x^2 - 3x + 2)`

`color(blue)(rArr -x^2 + 2x -1 - x^2 - 3x +2)`

`color(blue)(rArr -2x^2 -x +1 )`

Now we have

`y = -(2x^2 + x - 1)`

The General Form of a Quadratic Equation is

`color(red)(a)x^2 + color(red)(b)x + color(red)(c) = 0`

The Quadratic Formula for finding the roots of a quadratic equation is given by

`color(blue)(x=((-b)+- sqrt(b^2 - 4ac))/(2a),a!=0)`

`color(green)(b^2 - 4ac)` is referred to as the Discriminant

When `color(green)((b^2 - 4ac)=0)` there exists One Real Root

When `color(green)((b^2 - 4ac)>0)` there exists Two Real Roots

When `color(green)((b^2 - 4ac)<0)` there exists Two Complex Roots

For our Quadratic Equation

`-2x^2 - x + 1 = 0`

We observe that

`color(red)(a) = -2`

`color(red)(b) = -1`

`color(red)(c) = 1`

Substitute these values in the Quadratic Formula

`color(blue)(x=((-b)+- sqrt(b^2 - 4ac))/(2a)`

We get

`x=((-(-1))+- sqrt((-1)^2 - 4(-2)(1)))/(2(-2))`

On simplification we get,

`x=((+1)+- sqrt(+1 +8))/(-4)`

`x=(1+- sqrt(9))/(-4)`

We observe that, the discriminant `(b^2 - 4ac) > 0`

Hence, we will have Two Real Roots

`x = (1 +- 3)/-4`

`rArr x= (1+3)/-4; x = (1-3)/-4`

`rArr x= 4/-4; x = -2/-4`

`rArr x= -1; x = 1/2`

Hence our two real roots are

`color(blue)(x = 1/2 or x = 0.5)`

`color(blue)(x = -1)`

Hope this helps.