Question

How do you solve `x ^ { 2} + 10= 1`?

Answer 1

`x=3i, -3i`

Explanation:

`x^2+10=1`

Subtract 1 from both sides:

`x^2+10color(red)(-1)=1color(red)(-1)`

`x^2+9=0`

~~~~~

If we had a situation where we had `x^2-9=0`, we'd be able straight away say we could factor it to `(x+3)(x-3)`. But we don't and there is no easy way to find factors and so I'll use the Pythagorean Formula:

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

with `a=1, b=0, c=9`

`x = (0 \pm sqrt(0^2-4(1)(9))) / (2(1))`

`x = (pm sqrt(-36)) / 2`

Remember that `sqrt(-1)=i`, and so `sqrt(-36)=sqrt36xxsqrt(-1)=6i`

`:. x = (pm6i) / 2 =pm3i`