How do you solve 7d ^ { 2} + 10d + 168= 0?

Mar 25, 2018

$d = - \frac{5}{7} \pm \frac{\sqrt{1151}}{7} i$

Explanation:

We can find the complex roots by completing the square and using the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

with $A = 7 d + 5$ and $B = \sqrt{1151} i$ (where ${i}^{2} = - 1$) as follows:

$0 = 7 \left(7 {d}^{2} + 10 d + 168\right)$

$\textcolor{w h i t e}{0} = 49 {d}^{2} + 70 d + 1176$

$\textcolor{w h i t e}{0} = 49 {d}^{2} + 70 d + 25 + 1151$

$\textcolor{w h i t e}{0} = {\left(7 d\right)}^{2} + 2 \left(7 d\right) \left(5\right) + {\left(5\right)}^{2} + 1151$

$\textcolor{w h i t e}{0} = {\left(7 d + 5\right)}^{2} + {\left(\sqrt{1151}\right)}^{2}$

$\textcolor{w h i t e}{0} = {\left(7 d + 5\right)}^{2} - {\left(\sqrt{1151} i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(7 d + 5\right) - \sqrt{1151} i\right) \left(\left(7 d + 5\right) + \sqrt{1151} i\right)$

$\textcolor{w h i t e}{0} = \left(7 d + 5 - \sqrt{1151} i\right) \left(7 d + 5 + \sqrt{1151} i\right)$

So:

$7 d = - 5 \pm \sqrt{1151} i$

and:

$d = - \frac{5}{7} \pm \frac{\sqrt{1151}}{7} i$

Mar 25, 2018

$d = \frac{- 5 + i \sqrt{1151}}{7}$ or $d = \frac{- 5 - i \sqrt{1151}}{7}$

Explanation:

we use this formula

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

in this case our $a$ is $7$, $x$ is $d$, our $b$ is $10$ and our $c$ is $168$

putting them into equation gives us

$d = \frac{- 10 \pm \sqrt{{10}^{2} - \left(4\right) \left(7\right) \left(168\right)}}{2 \left(7\right)}$

by solving we get

$d = \frac{- 5 \pm i \sqrt{1151}}{7}$

$\implies d = \frac{- 5 - i \sqrt{1151}}{7} \mathmr{and} d = \frac{- 5 + i \sqrt{1151}}{7}$