How do you find the roots, real and imaginary, of #y=2x^2 -452x-68 # using the quadratic formula?

1 Answer
Dec 20, 2017

Answer:

Substitute in the values of #a#, #b# and #c# into the quadratic formula to get #x = 113 pm sqrt(12803)#.

Explanation:

We have #y = 2x^2 - 452x - 68#, where:

#a = 2#,

#b = -452#, and

#c = -68#.

Plug them in to the quadratic formula:

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

#x = (-(-452) pm sqrt((-452)^2 - 4(2)(-68)))/(2(2))#

We'll evaluate this starting from what's inside the square root. However, I think we should do some prime factorization.

#452 = 2^2 * 113#

#(-452)^2 = 2^4 * 113^2#

Substitute this back in:

#x = (-(-452) pm sqrt(2^4 * 113^2 - 4(2)(-68)))/(2(2))#

Then do the same for #-4(2)(-68)#:

#-4 * 2 * -68 = 4 * 2 * 68#

#4 * 2 * 68 = (2^2) * (2^1) * (2^2 * 17) = 2^5 * 17#

So we have:

#x = (-(-452) pm sqrt(2^4 * 113^2 + 2^5 * 17))/(2(2))#

Factor out what's common, which is #2^4#:

#x = (-(-452) pm sqrt(2^4(113^2 + 2 * 17)))/(2(2))#

Then evaluate #sqrt(2^4)#:

#x = (-(-452) pm 2^2 sqrt(113^2 + 2 * 17))/(2(2))#

#x = (-(-452) pm 4 sqrt(113^2 + 2 * 17))/(2(2))#

Now, let's look at what's outside the square root. #-(-452)# becomes #452#, and #2(2)# becomes #4#:

#x = (452 pm 4 sqrt(113^2 + 2 * 17))/4#

On the numerator, we can factor out #4#:

#x = (4(113 pm sqrt(113^2 + 2 * 17)))/4#

And they cancel out:

#x = 113 pm sqrt(113^2 + 2 * 17)#

Let's evaluate what's inside the square root:

#113^2 + 2 * 17 = 12769 + 2 * 17 = 12769 + 34 = 12803#

#x = 113 pm sqrt(12803)#

So we have two real solutions:

#x_1 = 113 + sqrt(12803)#

#x_2 = 113 - sqrt(12803)#

And no imaginary solutions. We could estimate the value of the square root to get:

#x_1 ~~ 113 + 113.15 = 226.15#

#x_2 ~~ 113 - 113.15 = -0.15#