How do you find the roots, real and imaginary, of $y= x^2 - x - (x-9)^2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y= x^2 - x - (x-9)^2$ using the quadratic formula?

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Answer 1 · 2017-03-19T07:46:46

We have real zero of the function, which is $81/17$

Explanation:

In the given function as we will see below terms relating to $x^2$ cancel out and hence, the function is a linear one and not quadratic and we do not need quadratic formula. Let us try to solve this.

$y=x^2-x-(x-9)^2$

= $x^2-x-(x^2-18x+81)$ - using formula for $(a-b)^2$

= $x^2-x-x^2+18x-81$

= $cancelx^2-x-cancelx^2+18x-81$

= $17x-81$

Hence, we have real zero of the function which is $81/17$

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How do you find the roots, real and imaginary, of $y= x^2 - x - (x-9)^2$ using the quadratic formula?

1 Answer

Explanation:

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Science

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Humanities

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How do you find the roots, real and imaginary, of y= x^2 - x - (x-9)^2 y= x^2 - x - (x-9)^2 using the quadratic formula?

1 Answer

Explanation:

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How do you find the roots, real and imaginary, of $y= x^2 - x - (x-9)^2$ using the quadratic formula?