How do you find the roots, real and imaginary, of $y = 2 {(x + 5)}^{2} + {(x - 4)}^{2}$ $y= 2(x+5)^2+(x-4)^2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = 2 {(x + 5)}^{2} + {(x - 4)}^{2}$ $y= 2(x+5)^2+(x-4)^2$ using the quadratic formula?

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Answer 1 · 2016-09-03T11:32:51

Zeros are $- 2 - 3 \sqrt{2} i$ $-2-3sqrt2i$ and $- 2 + 3 \sqrt{2} i$ $-2+3sqrt2i$

Explanation:

Quadratic formula gives the roots of $y = a x^{2} + b x + c$ $y=ax^2+bx+c$ as $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

As $y = 2 {(x + 5)}^{2} + {(x - 4)}^{2}$ $y=2(x+5)^2+(x-4)^2$

= $2 (x^{2} + 10 x + 25) + x^{2} - 8 x + 16$ $2(x^2+10x+25)+x^2-8x+16$

= $2 x^{2} + 20 x + 50 + x^{2} - 8 x + 16$ $2x^2+20x+50+x^2-8x+16$

= $3 x^{2} + 12 x + 66$ $3x^2+12x+66$

Hence zeros are given by

$x = \frac{- 12 \pm \sqrt{12^{2} - 4 \times 3 \times 66}}{2 \times 3}$ $x=(-12+-sqrt(12^2-4×3×66))/(2×3)$

= $\frac{- 12 \pm \sqrt{144 - 792}}{6}$ $(-12+-sqrt(144-792))/6$

= $\frac{- 12 \pm \sqrt{- 648}}{6}$ $(-12+-sqrt(-648))/6$

= $\frac{- 12 \pm 18 \sqrt{2} i}{6}$ $(-12+-18sqrt2i)/6$

= $- 2 \pm 3 \sqrt{2} i$ $-2+-3sqrt2i$

Hence, zeros are $- 2 - 3 \sqrt{2} i$ $-2-3sqrt2i$ and $- 2 + 3 \sqrt{2} i$ $-2+3sqrt2i$

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How do you find the roots, real and imaginary, of $y = 2 {(x + 5)}^{2} + {(x - 4)}^{2}$ $y= 2(x+5)^2+(x-4)^2$ using the quadratic formula?

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Explanation:

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How do you find the roots, real and imaginary, of $y = 2 {(x + 5)}^{2} + {(x - 4)}^{2}$ $y= 2(x+5)^2+(x-4)^2$ using the quadratic formula?