How do you find the roots, real and imaginary, of $y = {(x + 5)}^{2} + 6 x + 3$ $y=(x +5 )^2+6x+3$ using the quadratic formula?

Question

How do you find the roots, real and imaginary, of $y = {(x + 5)}^{2} + 6 x + 3$ $y=(x +5 )^2+6x+3$ using the quadratic formula?

1 Answer

Impact of this question

1499 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-07T02:07:07

$x 1 = - 14$ $x1 = -14$
$x 2 = - 2$ $x2 = -2$

Explanation:

$y = {(x + 5)}^{2} + 6 x + 3$ $y = (x + 5)^2 +6x +3$
Expand square term: $y = x^{2} + 10 x + 25 + 6 x + 3$ $y = x^2 + 10x + 25 + 6x + 3$
Therefore $y = x^{2} + 16 x + 28$ $y = x^2 +16x + 28$

This quadratic factorizes so there is no need of the Quadratic Formula:

$y = (x + 14) (x + 2)$ $y = (x + 14)(x + 2)$ which has zeros at $x = - 14 and - 2$ $x = -14 and -2$

If you would like to use the Quadratic Formula:

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

In this case $a = 1, b = 16, c = 28$ $a = 1, b = 16, c = 28$

Hence; $x = \frac{- 16 \pm \sqrt{256 - 112}}{2}$ $x = (-16 +- sqrt(256 - 112))/2$
$x = \frac{- 16 \pm \sqrt{144}}{2}$ $x = (-16+- sqrt(144))/2$
$x = \frac{- 16 \pm 12}{2}$ $x= (-16 +- 12)/2$

$x = - 14 or - 2$ $x = -14 or -2$

Regarding your question on real or complex roots, the roots of a quadratic function with real coefficients will either both be real or both be complex. In this case the roots are real.

Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of $y = {(x + 5)}^{2} + 6 x + 3$ $y=(x +5 )^2+6x+3$ using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of y=(x+5)2+6x+3y=(x +5 )^2+6x+3 using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the roots, real and imaginary, of $y = {(x + 5)}^{2} + 6 x + 3$ $y=(x +5 )^2+6x+3$ using the quadratic formula?