Question

Quadratic equations and complex numbers PLEASE HELPPPP ASAP

Answer 1

we are given

$27x^3-1=0$

we can also write it as

$(3x)^3-(1)^3=0$

now, we can use factor formula

$a^3-b^3=(a-b)(a^2+ab+b^2)$

we can use above formula

we get  

$(3x)^3-(1)^3=(3x-1)((3x)^2+3x\times 1+(1)^2)$

now, we can simplify it

$27x^3-1=(3x-1)(9x^2+3x+1)$

now, we can set it to 0

$27x^3-1=(3x-1)(9x^2+3x+1)=0$

and then we can solve for x

$(3x-1)(9x^2+3x+1)=0$

$3x-1=0$

$x=\frac{1}{3}$

$9x^2+3x+1=0$

now, we can use quadratic formula

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

now, we can plug values

and we get

$x=\frac{-3\pm \sqrt{3^2-4\cdot \:9\cdot \:1}}{2\cdot \:9}$

$x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}$

So, we will get solution as

$x=\frac{1}{3},\:x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}$...............Answer