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

Question

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

Avery Alexander · Accepted Answer

#x=(2isqrt3)/3, -(2isqrt3)/3#

We can first expand the bracket:

#y=2x^2+4x+4(x/2-1)^2#

#y=2x^2+4x+4(x^2/4-x/2-x/2+1)#

and now distribute the 4:

#y=2x^2+4x+4(x^2/4-x+1)#

#y=2x^2+4x+x^2-4x+4#

combine terms:

#y=3x^2+4#

and now see that we can use the quadratic formula, with values #a=3, b=0, c=4#

# x = (-b \pm sqrt(b^2-4ac)) / (2a) #

# x = (0 \pm sqrt(0^2-4(3)(4))) / (2(3)) #

# x = ( \pm sqrt(-48)) / 6 #

# x = ( \pm4i sqrt(3)) / 6 =pm(2isqrt3)/3#

Hazel Anglin · Answer

undefined To find the roots of $y = 2x^2 + 4x + 4\left(\frac{x}{2} - 1\right)^2$ using the quadratic formula, first, express the equation in the form $ax^2 + bx + c$. Then, apply the quadratic formula, which is:

$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

By comparing the given equation $y = 2x^2 + 4x + 4\left(\frac{x}{2} - 1\right)^2$ with $ax^2 + bx + c$, we can see that $a = 2$, $b = 4$, and $c = 4$.

Substitute these values into the quadratic formula:

$$x = \frac{{-4 \pm \sqrt{{4^2 - 4(2)(4)}}}}{{2(2)}}$$

$$x = \frac{{-4 \pm \sqrt{{16 - 32}}}}{{4}}$$

$$x = \frac{{-4 \pm \sqrt{{-16}}}}{{4}}$$

Since the discriminant ($b^2 - 4ac$) is negative, the roots will be imaginary. The square root of a negative number results in a complex number. Therefore, the roots of the given equation are imaginary.