How do you find the roots, real and imaginary, of $y = 2 x^{2} + 7 x + {(2 x - 1)}^{2}$ $y= 2x^2 + 7x+(2x- 1 )^2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = 2 x^{2} + 7 x + {(2 x - 1)}^{2}$ $y= 2x^2 + 7x+(2x- 1 )^2$ using the quadratic formula?

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Answer 1 · 2017-07-02T03:50:03

$x = \frac{- 3 + i \sqrt{15}}{12},$ $x=(-3+isqrt15)/12,$ $\frac{- 3 - i \sqrt{15}}{12}$ $(-3-isqrt15)/12$

Explanation:

Solve:

$y = 2 x^{2} + 7 x + {(2 x - 1)}^{2}$ $y=2x^2+7x+(2x-1)^2$

Expand ${(2 x - 1)}^{2}$ $(2x-1)^2$ using the square of a difference: $(a^{2} - b^{2}) = a^{2} - 2 a b + b^{2}$ $(a^2-b^2)=a^2-2ab+b^2$ , where $a = 2 x$ $a=2x$ and $b = 1$ $b=1$ .

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

Simplify.

${(2 x - 1)}^{2} = 4 x^{2} - 4 x + 1$ $(2x-1)^2=4x^2-4x+1$

Add the result to the rest of the equation.

$y = 2 x^{2} + 7 x + 4 x^{2} - 4 x + 1$ $y=2x^2+7x+4x^2-4x+1$

Simplify.

$y = 2 x^{2} + 4 x^{2} + 7 x - 4 x + 1$ $y=2x^2+4x^2+7x-4x+1$

$y = 6 x^{2} + 3 x + 1$ $y=6x^2+3x+1$

Substitute $0$ $0$ for $y$ $y$ .

$0 = 6 x^{2} + 3 x + 1$ $0=6x^2+3x+1$ is a quadratic equation in standard form: $a^{2} + b x + c$ $a^2+bx+c$ , where $a = 6$ $a=6$ , $b = 3$ $b=3$ , and $c = 1$ $c=1$ .

Use the quadratic formula to solve for the roots (values for $x$ $x$ ).

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

Substitute the values for $a, b, and c$ $a,b,andc$ into the formula.

$x = \frac{- 3 \pm \sqrt{3^{2} - 4 \cdot 6 \cdot 1}}{2 \cdot 6}$ $x=(-3+-sqrt(3^2-4*6*1))/(2*6)$

Simplify.

$x = \frac{- 3 \pm \sqrt{9 - 24}}{12}$ $x=(-3+-sqrt(9-24))/12$

$x = \frac{- 3 \pm \sqrt{- 15}}{12}$ $x=(-3+-sqrt(-15))/12$

$x = \frac{- 3 \pm i \sqrt{15}}{12}$ $x=(-3+-isqrt15)/12$

Roots:

$x = \frac{- 3 + i \sqrt{15}}{12},$ $x=(-3+isqrt15)/12,$ $\frac{- 3 - i \sqrt{15}}{12}$ $(-3-isqrt15)/12$

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How do you find the roots, real and imaginary, of $y = 2 x^{2} + 7 x + {(2 x - 1)}^{2}$ $y= 2x^2 + 7x+(2x- 1 )^2$ using the quadratic formula?

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Explanation:

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How do you find the roots, real and imaginary, of y=2x2+7x+(2x−1)2y= 2x^2 + 7x+(2x- 1 )^2 using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = 2 x^{2} + 7 x + {(2 x - 1)}^{2}$ $y= 2x^2 + 7x+(2x- 1 )^2$ using the quadratic formula?