How do you find the roots, real and imaginary, of $y = x^{2} + 8 x - 9 - {(- x - 1)}^{2}$ $y= x^2 + 8x -9- (-x-1)^2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = x^{2} + 8 x - 9 - {(- x - 1)}^{2}$ $y= x^2 + 8x -9- (-x-1)^2$ using the quadratic formula?

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Answer 1 · 2016-01-10T00:59:46

x=5/3

Explanation:

There are no imaginary roots of this equation.
To get the roots, put y=0
Now expand ${(- x - 1)}^{2}$ $(-x-1)^2$ which is actually ${(x + 1)}^{2}$ $(x+1)^2$

$x^{2} + 8 x - 9 - (x^{2} + 2 x + 1) = 0$ $x^2+8x-9-(x^2+2x+1)=0$

$cancel(x^2)+8x-9cancel(-x^2)-2x-1=0$

$6x-10=0$

$x=5/3$

Additionally you can also see that this equation is actually equation of a straight line so it can have at most 1 zero (which is real).

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How do you find the roots, real and imaginary, of $y = x^{2} + 8 x - 9 - {(- x - 1)}^{2}$ $y= x^2 + 8x -9- (-x-1)^2$ using the quadratic formula?

1 Answer

Explanation:

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How do you find the roots, real and imaginary, of y= x^2 + 8x -9- (-x-1)^2 y=x2+8x−9−(−x−1)2y= x^2 + 8x -9- (-x-1)^2 using the quadratic formula?

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Explanation:

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How do you find the roots, real and imaginary, of $y = x^{2} + 8 x - 9 - {(- x - 1)}^{2}$ $y= x^2 + 8x -9- (-x-1)^2$ using the quadratic formula?