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

Question

Nathan Axel · Accepted Answer

Form a quadratic equation, and solve it. Method below first, solution underneath.

Method Start by writing your equation in the form: #y=ax^2+bx+c# The roots are when #y=0# So you will have a quadratic equation of the form: #ax^2+bx+c=0# Then use the quadratic formula: #x=(-bpmsqrt(b^2-4ac))/(2a)# Which can give two roots due to the #pm#square root. Solution #y=-x^2+0x+2# for roots, #y=0# #-x^2+0x+2=0# #a=-1, b=0, c=2# so #x=(pmsqrt(8))/(-2)=pmsqrt2# #x=sqrt2# or #x=-sqrt2#

Abigail Allen · Answer

undefined To find the roots of $ y = -2(x+1)^2 + (-x-2)^2 $ using the quadratic formula, first, rewrite the equation in the standard quadratic form $ ax^2 + bx + c = 0 $. Then, identify $ a $, $ b $, and $ c $, and apply the quadratic formula $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $.

Expanding the given equation, we get $ y = -2x^2 - 4x - 2x - 4 $. Simplifying, we have $ y = -2x^2 - 6x - 4 $.

Comparing with the standard quadratic form, we have $ a = -2 $, $ b = -6 $, and $ c = -4 $.

Now, plug these values into the quadratic formula:

$$ x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4(-2)(-4)}}}}{{2(-2)}} $$

$$ x = \frac{{6 \pm \sqrt{{36 - 32}}}}{{-4}} $$

$$ x = \frac{{6 \pm \sqrt{4}}}{{-4}} $$

$$ x = \frac{{6 \pm 2}}{{-4}} $$

Thus, the roots are:

$$ x = \frac{4}{-4} = -1 $$

$$ x = \frac{8}{-4} = -2 $$

Therefore, the real roots are $ x = -1 $ and $ x = -2 $. There are no imaginary roots.