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#