How do you find the roots, real and imaginary, of $y= 5x^2 - 2x-45$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y= 5x^2 - 2x-45$ using the quadratic formula?

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Answer 1 · 2016-01-30T05:40:51

Since $b^2-4ac$ is 904 which is a positive number, the given quadratic equation has real roots.

$x=(2+-sqrt904)/10$

Explanation:

When the quadratic equation is in the form -

$ax^2+bx+c=0$

Then its roots are given by

$x=(-b+-sqrt(b^2-(4ac)))/(2a)$

In this $b^2-4ac$ determines whether a given equation has real roots or imaginary roots.

If $b^2-4ac$ is positive the roots are positive.

If $b^2-4ac$ is negative the roots are negative.

In our case -

$(-2)^2-(4xx5xx-45)$
$4 - (-900)$
$4+900=904$

Since $b^2-4ac$ is 904 which is a positive number, the given quadratic equation has real roots. So we should be able to plug into the quadratic formula without having imaginary numbers:

$x=(2+-sqrt(904))/(2(5))$

$x=(2+-sqrt904)/10$

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How do you find the roots, real and imaginary, of $y= 5x^2 - 2x-45$ using the quadratic formula?

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Explanation:

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How do you find the roots, real and imaginary, of y= 5x^2 - 2x-45 y= 5x^2 - 2x-45 using the quadratic formula?

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How do you find the roots, real and imaginary, of $y= 5x^2 - 2x-45$ using the quadratic formula?