How do you find the roots, real and imaginary, of $y = {(x + 2)}^{2} - 4 x + 14$ $y=(x +2 )^2-4x+14$ using the quadratic formula?

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Answer 1 · 2018-02-11T17:34:07

Roots are $- i 3 \sqrt{2}$ $-i3sqrt2$ and $i 3 \sqrt{2}$ $i3sqrt2$

$y = {(x + 2)}^{2} - 4 x + 14$ $y=(x+2)^2-4x+14$

= $x^{2} + 4 x + 4 - 4 x + 14$ $x^2+4x+4-4x+14$

= $x^{2} + 18$ $x^2+18$

= $x^{2} - {(i \sqrt{18})}^{2}$ $x^2-(isqrt18)^2$

= $(x + i \sqrt{18}) (x - i \sqrt{18})$ $(x+isqrt18)(x-isqrt18)$

= $(x + i 3 \sqrt{2}) (x - i 3 \sqrt{2})$ $(x+i3sqrt2)(x-i3sqrt2)$

Hence roots are $- i 3 \sqrt{2}$ $-i3sqrt2$ and $i 3 \sqrt{2}$ $i3sqrt2$

How do you find the roots, real and imaginary, of y=(x +2 )^2-4x+14y=(x+2)2−4x+14y=(x +2 )^2-4x+14 using the quadratic formula?