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

1 Answer
Dec 31, 2017

Answer:

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.