Question

When $x=\frac{3+5\sqrt{-1}}{2}$, find the value of $2x^3+2x^2-7x+72$; and show that it will be unaltered if $\frac{3-5\sqrt{-1}}{2}$ be substituted for $x$.

Answer 1

Let us form the equation whose roots are $\frac{3+5\sqrt{-1}}{2},\frac{3-5\sqrt{-1}}{2}$

Sum of the roots is $3$ and the product of the roots is $\frac{17}{2}$

So the required equation is $2x^2-6x+17=0$

Therefore for $x=\frac{3+5\sqrt{-1}}{2},\frac{3-5\sqrt{-1}}{2}$ , the equation $2x^2-6x+17$ will vanish.

Now the equation $2x^3+2x^2-7x+72$ can be written as $x(2x^2-6x+17)+4(2x^2-6x+17)+4$

Therefore the value of $2x^3+2x^2-7x+72$ at $x=\frac{3+5\sqrt{-1}}{2},\frac{3-5\sqrt{-1}}{2}$ is $x\left(0\right)+4\left(0\right)+4=4$