Question

How do you find the roots, real and imaginary, of `y= 2x^2 + 7x+(2x- 1 )^2` using the quadratic formula?

Answer 1

`x=(-3+isqrt15)/12,` `(-3-isqrt15)/12`

Explanation:

Solve:

`y=2x^2+7x+(2x-1)^2`

Expand `(2x-1)^2` using the square of a difference: `(a^2-b^2)=a^2-2ab+b^2`, where `a=2x` and `b=1`.

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

Simplify.

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

Add the result to the rest of the equation.

`y=2x^2+7x+4x^2-4x+1`

Simplify.

`y=2x^2+4x^2+7x-4x+1`

`y=6x^2+3x+1`

Substitute `0` for `y`.

`0=6x^2+3x+1` is a quadratic equation in standard form: `a^2+bx+c`, where `a=6`, `b=3`, and `c=1`.

Use the quadratic formula to solve for the roots (values for `x`).

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

Substitute the values for `a,b,andc` into the formula.

`x=(-3+-sqrt(3^2-4*6*1))/(2*6)`

Simplify.

`x=(-3+-sqrt(9-24))/12`

`x=(-3+-sqrt(-15))/12`

`x=(-3+-isqrt15)/12`

Roots:

`x=(-3+isqrt15)/12,` `(-3-isqrt15)/12`