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

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Answer 1 · 2016-02-02T15:01:37

We have two real roots: $x = - 1 +- sqrt(2)/2$ .

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

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

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

... use the formula $(m-n)^2 = m^2 - 2mn + n^2$ ....

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

$color(white)(x) = - 2x^2 - 4x - 1$

Now, we are ready to use the quadratic formula.

The formula is

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

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

$x = (4 +- sqrt((-4)^2 - 4 * (-2) * (-1)))/(2 * (-2)) = (4 +- sqrt(16 - 8))/(-4) = (4 +- sqrt(8))/(-4) = (4 +- 2 sqrt(2))/(-4) = -1 +- sqrt(2)/2$

So, we have two real solutions:

$x_1 = -1 + sqrt(2)/2 ~~ - 0.29$

and

$x_2 = -1 - sqrt(2)/2 ~~ -1.71$

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