How do you find the roots, real and imaginary, of $y = x^{2} - 39 x - 2 {(3 x - 1)}^{2}$ $y= x^2-39x-2(3x-1)^2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = x^{2} - 39 x - 2 {(3 x - 1)}^{2}$ $y= x^2-39x-2(3x-1)^2$ using the quadratic formula?

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Answer 1 · 2017-07-17T15:38:14

$0.04 or 2.644$ $0.04 " or " 2.644$

Explanation:

Right now, you have $y = x^{2} - 39 x - 2 {(3 x - 1)}^{2}$ $y=x^2-39x-2(3x-1)^2$

First, we need to expand ${(3 x - 1)}^{2}$ $(3x-1)^2$ .

To expand the brackets in the form of ${(a x + b)}^{2}$ $(color(red)(a)color(green)(x)+color(blue)(b))^2$ , we do ${(a x)}^{2} + 2 (a x b) + b^{2}$ $(color(red)(a)color(green)(x))^2+2(color(red)(a)color(green)(x)color(blue)(b))+color(blue)(b)^2$

In this case we have ${(3 x + - 1)}^{2}$ $(color(red)(3)color(green)(x)+color(blue)(-1))^2$

${(3 x)}^{2} = 9 x^{2}$ $(3x)^2=9x^2$
$2 \cdot 3 x \cdot - 1 = - 6 x$ $2*3x*-1=-6x$
$- 1^{2} = 1$ $-1^2=1$
$9 x^{2} - 6 x + 1$ $9x^2-6x+1$

But we need 2 lots, so $2 (9 x^{2} - 6 x + 1) = 18^{2} - 12 x + 2$ $2(9x^2-6x+1)=18^2-12x+2$

$y = x^{2} - 39 x + 18 x^{2} - 12 x + 2 = 19 x^{2} - 51 x + 2$ $y=x^2-39x+18x^2-12x+2=19x^2-51x+2$

The quadratic formula is $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \equiv \frac{- - 51 \pm \sqrt{{- 51}^{2} - 4 (19 \cdot 4)}}{2 (19)} = \frac{51 \pm \sqrt{2601 - 152}}{38} = 0.04 or 2.644$ $x=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4color(red)(a)color(orange)(c)))/(2color(red)(a))-=(-color(blue)(-51)+-sqrt(color(blue)(-51)^2-4(color(red)(19)*color(orange)(4))))/(2(color(red)(19)))=(color(blue)(51)+-sqrt(color(blue)2601-color(yellow)(152)))/(38)=0.04 " or " 2.644$

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How do you find the roots, real and imaginary, of $y = x^{2} - 39 x - 2 {(3 x - 1)}^{2}$ $y= x^2-39x-2(3x-1)^2$ using the quadratic formula?

1 Answer

Explanation:

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How do you find the roots, real and imaginary, of y=x2−39x−2(3x−1)2y= x^2-39x-2(3x-1)^2 using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = x^{2} - 39 x - 2 {(3 x - 1)}^{2}$ $y= x^2-39x-2(3x-1)^2$ using the quadratic formula?