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

1 Answer
Dec 20, 2017

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