How do you find the roots, real and imaginary, of $y=4x^2 + -7x -(x-2)^2$ using the quadratic formula?

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Answer 1 · 2017-08-23T04:18:17

$x=(3+sqrt57)/6,$ $(3-sqrt57)/6$

Solve:

$y=4x^2+(-7x)-(x-2)^2$

Expand $(x-2)^2$ .

$y=4x^2+(-7x)-(x-2)(x-2)$

$y=4x^2+(-7x)-(x^2-4x+4)$

Simplify.

$y=4x^2-7x-x^2+4x-4$

Gather like terms.

$y=(4x^2+(-x^2))+(-7x+4x)-4$

Combine like terms.

$y=3x^2-3x-4$

To solve for roots, substitute $0$ for $y$ . Then solve for $x$ using the quadratic formula.

$0=3x^2-3x-4$

Quadratic formula

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

Plug in the given values.

$x=(-(-3)+-sqrt((-3)^2-4*3*-4))/(2*3)$

Simplify.

$x=(3+-sqrt(9+48))/6$

$x=(3+-sqrt57)/12$ $larr$ $57$ is the product of two prime numbers.

$x=(3+sqrt57)/6,$ $(3-sqrt57)/6$

How do you find the roots, real and imaginary, of y=4x^2 + -7x -(x-2)^2 y=4x^2 + -7x -(x-2)^2 using the quadratic formula?