How do you find the roots, real and imaginary, of y= 2 x(x - 4) -(2x-1)^2 y=2x(x−4)−(2x−1)2 using the quadratic formula?

1 Answer
Lotusbluete
Feb 2, 2016

We have two real roots: x = - 1 +- sqrt(2)/2x=−1±√22.

Explanation:

First of all, let's expand and simplify the expression.

y = 2x(x-4) - (2x-1)^2 y=2x(x−4)−(2x−1)2

color(white)(x) = 2x^2 - 8x - (4x^2 - 4x + 1)x=2x2−8x−(4x2−4x+1)

... use the formula (m-n)^2 = m^2 - 2mn + n^2(m−n)2=m2−2mn+n2 ....

color(white)(x) = 2x^2 - 8x - 4x^2 + 4x - 1x=2x2−8x−4x2+4x−1

color(white)(x) = - 2x^2 - 4x - 1x=−2x2−4x−1

Now, we are ready to use the quadratic formula.

The formula is

x = (-b +- sqrt(b^2 - 4ac))/(2a)x=−b±√b2−4ac2a

In our case, a = -2a=−2, b = -4b=−4 and c = -1c=−1. So, we have

x = (4 +- sqrt((-4)^2 - 4 * (-2) * (-1)))/(2 * (-2)) = (4 +- sqrt(16 - 8))/(-4) = (4 +- sqrt(8))/(-4) = (4 +- 2 sqrt(2))/(-4) = -1 +- sqrt(2)/2x=4±√(−4)2−4⋅(−2)⋅(−1)2⋅(−2)=4±√16−8−4=4±√8−4=4±2√2−4=−1±√22

So, we have two real solutions:

x_1 = -1 + sqrt(2)/2 ~~ - 0.29x1=−1+√22≈−0.29

and

x_2 = -1 - sqrt(2)/2 ~~ -1.71x2=−1−√22≈−1.71

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