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

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Answer 1 · 2017-07-18T02:03:44

See a solution process below:

First, we need to expand the squared term on the right side of the equation using the rule:

#(a + b)^2 = a^2 + 2ab + b^2

Substituting $x$ $x$ for $a$ $a$ and $2$ $2$ for $b$ $b$ gives:

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

$y = x^{2} + (2 \times x \times 2) + 2^{2} - x - 5$ $y = x^2 + (2 xx x xx 2) + 2^2 - x - 5$

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

We can now group and combine like terms to put the equation in standard form for a quadratic.

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

$y = x^{2} + 4 x - 1 x + 4 - 5$ $y = x^2 + 4x - 1x + 4 - 5$

$y = x^{2} + (4 - 1) x + (4 - 5)$ $y = x^2 + (4 - 1)x + (4 - 5)$

$y = x^{2} + 3 x + (- 1)$ $y = x^2 + 3x + (-1)$

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

The quadratic formula states:

For $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ , the values of $x$ $x$ which are the solutions to the equation are given by:

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

Substituting $1$ $1$ for $a$ $a$ ; $3$ $3$ for $b$ $b$ and $- 1$ $-1$ for $c$ $c$ gives:

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

$x = \frac{- 3 \pm \sqrt{9 - (- 4)}}{2}$ $x = (-3 +- sqrt(9 - (-4)))/2$

$x = \frac{- 3 \pm \sqrt{9 + 4}}{2}$ $x = (-3 +- sqrt(9 + 4))/2$

$x = \frac{- 3 \pm \sqrt{13}}{2}$ $x = (-3 +- sqrt(13))/2$

Or

$x = - \frac{3}{2} + \frac{\sqrt{13}}{2}$ $x = -3/2 + sqrt(13)/2$ and $x = - \frac{3}{2} - \frac{\sqrt{13}}{2}$ $x = -3/2 - sqrt(13)/2$

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