Question

If $\alpha$ & $\beta^2$ are the roots of the equation, $8x^2 - 10x + 3 =$ $\frac{1}{0}$, wehre $\beta^2 > \frac{1}{2}$, then the equation whose roots are $(\alpha$ $+i\beta)^{100}$ & $(\alpha-i\beta)^{100}$ is :

• A. $x^2-x+1=0$
• B. $x^2+x+1=0$
• C. $x^2-x-1=0$
• D. $x^2+x-1=0$

Answer 1

Answer: (B)

x^2+x+1=0

Answer 2

Solution:

1. The quadratic equation is $8x^2 - 10x + 3 = 0.$ Its discriminant $D = 10^2 - 4 \cdot 8 \cdot 3 = 100 - 96 = 4$. Therefore, the roots are $\frac{10 \pm 2}{16} = \frac{12}{16} = \frac{3}{4} \quad \text{and} \quad \frac{8}{16} = \frac{1}{2}.$

2. Given that one root is $\beta^2$ and $\beta^2 > \frac{1}{2}$, we have $\beta^2 = \frac{3}{4}$ and so $\alpha = \frac{1}{2}$. Then, $\beta = \frac{\sqrt{3}}{2}$ (taking the positive value).

3. Now, $\alpha + i\beta = \frac{1}{2} + i\frac{\sqrt{3}}{2} = e^{i\pi/3},$ $\alpha - i\beta = \frac{1}{2} - i\frac{\sqrt{3}}{2} = e^{-i\pi/3}.$

4. Raising these to the 100th power: $(\alpha + i\beta)^{100} = e^{i(100\pi/3)}$ $(\alpha - i\beta)^{100} = e^{-i(100\pi/3)}.$

5. Reduce the exponent modulo $2\pi$:

   For positive exponent: $\frac{100\pi}{3} - 32\pi = \frac{100\pi - 96\pi}{3} = \frac{4\pi}{3}$. Hence, one root is $e^{i4\pi/3} = -\tfrac{1}{2} - i\frac{\sqrt{3}}{2}$.

   For negative exponent: $-\frac{100\pi}{3} + 34\pi = \frac{-100\pi + 102\pi}{3} = \frac{2\pi}{3}$. Hence, the other root is $e^{i2\pi/3} = -\tfrac{1}{2} + i\frac{\sqrt{3}}{2}$.

6. Their sum and product: Sum $= e^{i4\pi/3}+e^{i2\pi/3} = \left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)+\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = -1.$ Product $= e^{i(4\pi/3+2\pi/3)} = e^{i2\pi} = 1.$

7. The quadratic equation with roots $r_1$ and $r_2$ is: $x^2 - (r_1 + r_2)x + r_1r_2 = x^2 - (-1)x + 1 = x^2 + x + 1 = 0.$

Explanation (minimal): Roots of $8x^2-10x+3=0$ are $\frac{3}{4}$ and $\frac{1}{2}$. Given $\beta^2>\frac{1}{2}$, take $\beta^2=\frac{3}{4}$ and $\alpha=\frac{1}{2}$. Then, $\alpha \pm i\beta=e^{\pm i\pi/3}$ and raised to the 100th power yield $e^{i4\pi/3}$ and $e^{i2\pi/3}$. Their sum is $-1$ and product is $1$, forming the quadratic $x^2+x+1=0$.