How do you find the roots, real and imaginary, of #y= x^2 - 8x + (x+4)^2 # using the quadratic formula?

Question

Hazel Anglin · Accepted Answer

#color(maroon)("two roots are " +2sqrt2 i, -2sqrt2 i# #y = x^2 - 8x + (x + 4)^2# #y = x^2 - 8x + x^2 + 8x + 16# #y = 2x^2 + 16# #x = (-b +- sqrt(b^2 - 4ac)) / (2a)# #a = 2, b = 0, c = 16# #x = (0 +- sqrt(0 - 128)) / 4# #x = +- 8sqrt(-2) / 4 = +- 2 sqrt(-2)# #color(maroon)(x = + 2sqrt2 i, -2sqrt2 i#

Hazel Anglin · Answer

undefined To find the roots of the quadratic equation $ y = x^2 - 8x + (x+4)^2 $ using the quadratic formula, we first need to rewrite the equation in standard form. Then, we can identify the coefficients $ a $, $ b $, and $ c $ in the form $ ax^2 + bx + c $, and substitute them into the quadratic formula $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $.

Given the equation $ y = x^2 - 8x + (x+4)^2 $, we can expand $ (x+4)^2 $ to get:

$$ y = x^2 - 8x + (x^2 + 8x + 16) = 2x^2 + 16 $$

Now, comparing with the standard form $ ax^2 + bx + c $, we have $ a = 2 $, $ b = 0 $, and $ c = 16 $.

Substituting these values into the quadratic formula, we get:

$$ x = \frac{{-0 \pm \sqrt{{0^2 - 4 \cdot 2 \cdot 16}}}}{{2 \cdot 2}} $$

$$ x = \frac{{\pm \sqrt{{-128}}}}{{4}} $$

$$ x = \frac{{\pm \sqrt{{128}}i}}{{4}} $$

$$ x = \frac{{\pm 8i}}{{4}} $$

$$ x = \pm 2i $$

So, the roots of the equation are $ x = 2i $ and $ x = -2i $, which are imaginary roots.