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

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Answer 1 · 2018-07-14T20:17:19

$x = - 8 \pm \sqrt{145}$ $x=-8pmsqrt(145)$

Simplifying the given equation and using the Formula
${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a-b)^2=a^2-2ab+b^2$

we get
$2 x^{2} - 2 x - (x^{2} - 18 x + 81) = 0$ $2x^2-2x-(x^2-18x+81)=0$ combinging like Terms

$x^{2} + 16 x - 81 = 0$ $x^2+16x-81=0$

$x^{2} + p x + q = 0$ $x^2+px+q=0$

$x_{1, 2} = - \frac{p}{2} \pm \sqrt{{(\frac{p}{2})}^{2} - q}$ $x_(1,2)=-p/2pmsqrt((p/2)^2-q)$

we get

$x_{1, 2} = - 8 \pm \sqrt{64 + 81}$ $x_(1,2)=-8pmsqrt(64+81)$

$x_{1, 2} = - 8 \pm \sqrt{145}$ $x_(1,2)=-8pmsqrt(145)$

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