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

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

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Answer 1 · 2017-12-20T07:40:29

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

Explanation:

We have $y = 2 x^{2} - 452 x - 68$ $y = 2x^2 - 452x - 68$ , where:

$a = 2$ $a = 2$ ,

$b = - 452$ $b = -452$ , and

$c = - 68$ $c = -68$ .

Plug them in to the quadratic formula:

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

$x = \frac{- (- 452) \pm \sqrt{{(- 452)}^{2} - 4 (2) (- 68)}}{2 (2)}$ $x = (-(-452) pm sqrt((-452)^2 - 4(2)(-68)))/(2(2))$

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

$452 = 2^{2} \cdot 113$ $452 = 2^2 * 113$

${(- 452)}^{2} = 2^{4} \cdot 113^{2}$ $(-452)^2 = 2^4 * 113^2$

Substitute this back in:

$x = \frac{- (- 452) \pm \sqrt{2^{4} \cdot 113^{2} - 4 (2) (- 68)}}{2 (2)}$ $x = (-(-452) pm sqrt(2^4 * 113^2 - 4(2)(-68)))/(2(2))$

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

$- 4 \cdot 2 \cdot - 68 = 4 \cdot 2 \cdot 68$ $-4 * 2 * -68 = 4 * 2 * 68$

$4 \cdot 2 \cdot 68 = (2^{2}) \cdot (2^{1}) \cdot (2^{2} \cdot 17) = 2^{5} \cdot 17$ $4 * 2 * 68 = (2^2) * (2^1) * (2^2 * 17) = 2^5 * 17$

So we have:

$x = \frac{- (- 452) \pm \sqrt{2^{4} \cdot 113^{2} + 2^{5} \cdot 17}}{2 (2)}$ $x = (-(-452) pm sqrt(2^4 * 113^2 + 2^5 * 17))/(2(2))$

Factor out what's common, which is $2^{4}$ $2^4$ :

$x = \frac{- (- 452) \pm \sqrt{2^{4} (113^{2} + 2 \cdot 17)}}{2 (2)}$ $x = (-(-452) pm sqrt(2^4(113^2 + 2 * 17)))/(2(2))$

Then evaluate $\sqrt{2^{4}}$ $sqrt(2^4)$ :

$x = \frac{- (- 452) \pm 2^{2} \sqrt{113^{2} + 2 \cdot 17}}{2 (2)}$ $x = (-(-452) pm 2^2 sqrt(113^2 + 2 * 17))/(2(2))$

$x = \frac{- (- 452) \pm 4 \sqrt{113^{2} + 2 \cdot 17}}{2 (2)}$ $x = (-(-452) pm 4 sqrt(113^2 + 2 * 17))/(2(2))$

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

$x = \frac{452 \pm 4 \sqrt{113^{2} + 2 \cdot 17}}{4}$ $x = (452 pm 4 sqrt(113^2 + 2 * 17))/4$

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

$x = \frac{4 (113 \pm \sqrt{113^{2} + 2 \cdot 17})}{4}$ $x = (4(113 pm sqrt(113^2 + 2 * 17)))/4$

And they cancel out:

$x = 113 \pm \sqrt{113^{2} + 2 \cdot 17}$ $x = 113 pm sqrt(113^2 + 2 * 17)$

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

$113^{2} + 2 \cdot 17 = 12769 + 2 \cdot 17 = 12769 + 34 = 12803$ $113^2 + 2 * 17 = 12769 + 2 * 17 = 12769 + 34 = 12803$

$x = 113 \pm \sqrt{12803}$ $x = 113 pm sqrt(12803)$

So we have two real solutions:

$x_{1} = 113 + \sqrt{12803}$ $x_1 = 113 + sqrt(12803)$

$x_{2} = 113 - \sqrt{12803}$ $x_2 = 113 - sqrt(12803)$

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

$x_{1} \approx 113 + 113.15 = 226.15$ $x_1 ~~ 113 + 113.15 = 226.15$

$x_{2} \approx 113 - 113.15 = - 0.15$ $x_2 ~~ 113 - 113.15 = -0.15$

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

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Explanation:

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