How do you find the roots, real and imaginary, of $y = x^{2} - 21 x + {(x + 1)}^{2}$ $y= x^2 -21x +(x+1)^2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = x^{2} - 21 x + {(x + 1)}^{2}$ $y= x^2 -21x +(x+1)^2$ using the quadratic formula?

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Answer 1 · 2017-08-10T07:14:17

$x = \frac{19 \pm \sqrt{353}}{4}$ $x = frac(19 pm sqrt(353))(4)$

Explanation:

We have: $y = x^{2} - 21 x + {(x + 1)}^{2}$ $y = x^(2) - 21 x + (x + 1)^(2)$

First, let's expand the parentheses:

$\Rightarrow y = x^{2} - 21 x + x^{2} + 2 x + 1$ $Rightarrow y = x^(2) - 21 x + x^(2) + 2 x + 1$

$\Rightarrow y = 2 x^{2} - 19 x + 1$ $Rightarrow y = 2 x^(2) - 19 x + 1$

Then, let's apply the quadratic formula:

$\Rightarrow x = \frac{- (- 19) \pm \sqrt{{(- 19)}^{2} - 4 (2) (1)}}{2 (2)}$ $Rightarrow x = frac(- (- 19) pm sqrt((- 19)^(2) - 4(2)(1)))(2(2))$

$\Rightarrow x = \frac{19 \pm \sqrt{361 - 8}}{4}$ $Rightarrow x = frac(19 pm sqrt(361 - 8))(4)$

$\Rightarrow x = \frac{19 \pm \sqrt{353}}{4}$ $Rightarrow x = frac(19 pm sqrt(353))(4)$

Therefore, the solutions to the equation are $x = \frac{19 - \sqrt{353}}{4}$ $x = frac(19 - sqrt(353))(4)$ and $x = \frac{19 + \sqrt{353}}{4}$ $x = frac(19 + sqrt(353))(4)$ .

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How do you find the roots, real and imaginary, of $y = x^{2} - 21 x + {(x + 1)}^{2}$ $y= x^2 -21x +(x+1)^2$ using the quadratic formula?

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Explanation:

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How do you find the roots, real and imaginary, of y= x^2 -21x +(x+1)^2 y=x2−21x+(x+1)2y= x^2 -21x +(x+1)^2 using the quadratic formula?

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How do you find the roots, real and imaginary, of $y = x^{2} - 21 x + {(x + 1)}^{2}$ $y= x^2 -21x +(x+1)^2$ using the quadratic formula?