Question

One of the root of the equation $4x^4-24x^3+57x^2+18x-45=0$ is $3+\sqrt{6}i$, $i=\sqrt{-1}$. The number of real roots of the equation is

Answer 1

$x=3+\sqrt{6}i$

$\Rightarrow x-3=\sqrt{6}i$

squaring both sides, we get

$\Rightarrow {(x-3)}^2={\left(\sqrt{6}i\right)}^2$

$\Rightarrow x^2-6x+9=6i^2=-6$ where $i^2=-1$

$\Rightarrow x^2-6x+9+6=0$

$\Rightarrow x^2-6x+15=0$ .........$\left(1\right)$

From the question

$4x^4-24x^3+57x^2+18x-45=0$

$\Rightarrow 4x^4-24x^3+(56x^2+x^2)+(-6x+24x)+(15-60)=0$

$\Rightarrow 4x^4-24x^3+56x^2+24x-60+(x^2-6x+15)=0$ from $\left(1\right)$

$\Rightarrow 4x^4-24x^3+56x^2+24x-60+0=0$ from $\left(1\right)$

$\Rightarrow 4x^4-24x^3+56x^2+24x-60=0$

$\Rightarrow x^4-6x^3+16x^2+6x-15=0$ by dividing the complete equation by $4$

$\Rightarrow x^4-6x^3+(13x^2+x^2)+(12x-6x)+(15-30)=0$

$\Rightarrow x^4-6x^3+13x^2+12x-30+(x^2-6x+15)=0$ from $\left(1\right)$

$\Rightarrow x^4-6x^3+13x^2+12x-30=0$

$\Rightarrow x^2(x^2-6x+13)+12x-30=0$

$\Rightarrow x^2(x^2-6x+15-2)+12x-30=0$

$\Rightarrow x^2(x^2-6x+15)-2x^2+12x-30=0$

$\Rightarrow x^2\times 0-2x^2+12x-30=0$ from $\left(1\right)$

$\Rightarrow -2x^2+12x-30=0$

$\Rightarrow x^2-6x+15=0$ by dividing by $-2$

$\Rightarrow$Discriminant$=b^2-4ac=6^2-4\times 1\times 15=36-60=-24<0$

$\therefore$ there are no real roots.