How do you find the roots, real and imaginary, of $y=-5(x-3)^2-(x-3)^2-10$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y=-5(x-3)^2-(x-3)^2-10$ using the quadratic formula?

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Answer 1 · 2016-04-26T12:04:15

$3-i10sqrt2$ and $3+i10sqrt2$ are roots of given equation.

Explanation:

For equation $ax^2+bx+c=0$ , quadratic formula gives $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

In equation $y=-5(x-3)^2-(x-3)^2-10=-6(x-3)^2-10$ . Hence, $(x-3)=(0+-sqrt(0^2-4(-5)(-10)))/(2xx(-5))=(+-sqrt(-200))/(-10)=+-i10sqrt2/(-10)=+-i10sqrt2$

Hence, $x=3+-i10sqrt2$ and these are roots of given equation.

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How do you find the roots, real and imaginary, of $y=-5(x-3)^2-(x-3)^2-10$ using the quadratic formula?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of y=-5(x-3)^2-(x-3)^2-10 y=-5(x-3)^2-(x-3)^2-10 using the quadratic formula?

1 Answer

Explanation:

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How do you find the roots, real and imaginary, of $y=-5(x-3)^2-(x-3)^2-10$ using the quadratic formula?