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

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Solution

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 $2 x^{2} - 6 x + 17 = 0$
Therefore for $x = \frac{3 + 5 \sqrt{- 1}}{2}, \frac{3 - 5 \sqrt{- 1}}{2}$ , the equation $2 x^{2} - 6 x + 17$ will vanish.
Now the equation $2 x^{3} + 2 x^{2} - 7 x + 72$ can be written as $x (2 x^{2} - 6 x + 17) + 4 (2 x^{2} - 6 x + 17) + 4$
Therefore the value of $2 x^{3} + 2 x^{2} - 7 x + 72$ at $x = \frac{3 + 5 \sqrt{- 1}}{2}, \frac{3 - 5 \sqrt{- 1}}{2}$ is $x (0) + 4 (0) + 4 = 4$

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