Question

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

Answer 1

`d = -5/7+-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 = (A-B)(A+B)`

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

`0 = 7(7d^2+10d+168)`

`color(white)(0) = 49d^2+70d+1176`

`color(white)(0) = 49d^2+70d+25+1151`

`color(white)(0) = (7d)^2+2(7d)(5)+(5)^2+1151`

`color(white)(0) = (7d+5)^2+(sqrt(1151))^2`

`color(white)(0) = (7d+5)^2-(sqrt(1151)i)^2`

`color(white)(0) = ((7d+5)-sqrt(1151)i)((7d+5)+sqrt(1151)i)`

`color(white)(0) = (7d+5-sqrt(1151)i)(7d+5+sqrt(1151)i)`

So:

`7d = -5+-sqrt(1151)i`

and:

`d = -5/7+-sqrt(1151)/7 i`

Answer 2

`d=(-5+isqrt1151)/7` or `d=(-5-isqrt1151)/7`

Explanation:

we use this formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

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=(-10+-sqrt(10^2-(4)(7)(168)))/(2(7))`

by solving we get

`d=(-5+-isqrt1151)/7`

`=>d=(-5-isqrt1151)/7 or d=(-5+isqrt1151)/7`