How do you find the roots, real and imaginary, of $y= -13x^2 -5x +11(x-3)^2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y= -13x^2 -5x +11(x-3)^2$ using the quadratic formula?

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Answer 1 · 2016-08-04T17:48:35

We have two irrational zeros $-71/4-1/4sqrt5833$ and $-71/4+1/4sqrt5833$

Explanation:

$y=-13x^2-5x+11(x-3)^2$

= $-13x^2-5x+11(x^2-6x+9)$

= $-13x^2-5x+11x^2-66x+99$

= $-2x^2-71x+99$

As discriminant $b^2-4ac=(-71)^2-4xx(-2)xx99$

= $5041+792=5833$ which is positive but not a square of a rational number and hence we have irrational numbers as zeros of te function $y=-13x^2-5x+11(x-3)^2$ and these are given by quadratic formula

$(-(-71)+-sqrt5833)/(-4)=-71/4+-1/4sqrt5833$ or

$-71/4-1/4sqrt5833$ and $-71/4+1/4sqrt5833$

Note: Only factors of $5833$ are $19$ and $307$ .

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How do you find the roots, real and imaginary, of $y= -13x^2 -5x +11(x-3)^2$ using the quadratic formula?

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Explanation:

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Science

Math

Humanities

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How do you find the roots, real and imaginary, of y= -13x^2 -5x +11(x-3)^2 y= -13x^2 -5x +11(x-3)^2 using the quadratic formula?

1 Answer

Explanation:

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How do you find the roots, real and imaginary, of $y= -13x^2 -5x +11(x-3)^2$ using the quadratic formula?