Question

If $\alpha$ and $\beta$ are the roots of the equation, $7x^2 - 3x - 2 = 0$, then the value of $\frac{\alpha}{1-\alpha^2} + \frac{\beta}{1-\beta^2}$ is equal to

• A. 3/8
• B. 1/24
• C. 27/16
• D. 27/32

Answer 1

Answer: (C)

27/16

Answer 2

The problem asks for the value of a symmetric expression involving the roots of a quadratic equation.

Given the quadratic equation: $7x^2 - 3x - 2 = 0$. Let $\alpha$ and $\beta$ be the roots of this equation. From Vieta's formulas, we have:

Sum of roots: $\alpha + \beta = -(\text{coefficient of } x) / (\text{coefficient of } x^2) = -(-3)/7 = 3/7$

Product of roots: $\alpha\beta = (\text{constant term}) / (\text{coefficient of } x^2) = -2/7$

We need to find the value of the expression $E = \frac{\alpha}{1-\alpha^2} + \frac{\beta}{1-\beta^2}$.

Method 1: Direct Simplification

Combine the terms by finding a common denominator: $E = \frac{\alpha(1-\beta^2) + \beta(1-\alpha^2)}{(1-\alpha^2)(1-\beta^2)}$

Expand the numerator: Numerator $= \alpha - \alpha\beta^2 + \beta - \beta\alpha^2 = (\alpha+\beta) - \alpha\beta(\beta+\alpha)$

Factor out $(\alpha+\beta)$: Numerator $= (\alpha+\beta)(1 - \alpha\beta)$

Expand the denominator: Denominator $= 1 - \beta^2 - \alpha^2 + \alpha^2\beta^2 = 1 - (\alpha^2+\beta^2) + (\alpha\beta)^2$

To find $\alpha^2+\beta^2$, use the identity $(\alpha+\beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$: $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$

Substitute the values of $\alpha+\beta$ and $\alpha\beta$: $\alpha^2+\beta^2 = (3/7)^2 - 2(-2/7)$ $\alpha^2+\beta^2 = 9/49 + 4/7$

To add these fractions, find a common denominator (49): $\alpha^2+\beta^2 = 9/49 + (4 \times 7)/(7 \times 7) = 9/49 + 28/49 = 37/49$

Now substitute these values back into the numerator and denominator expressions: Numerator $= (\alpha+\beta)(1 - \alpha\beta) = (3/7)(1 - (-2/7)) = (3/7)(1 + 2/7) = (3/7)(7/7 + 2/7) = (3/7)(9/7) = 27/49$

Denominator $= 1 - (\alpha^2+\beta^2) + (\alpha\beta)^2 = 1 - (37/49) + (-2/7)^2 = 1 - 37/49 + 4/49 = 49/49 - 37/49 + 4/49 = (49 - 37 + 4)/49 = (12 + 4)/49 = 16/49$

Finally, divide the numerator by the denominator: $E = \frac{27/49}{16/49} = \frac{27}{16}$

Method 2: Using the given equation

Since $\alpha$ is a root of $7x^2 - 3x - 2 = 0$, it satisfies the equation: $7\alpha^2 - 3\alpha - 2 = 0$

We can express $\alpha^2$ in terms of $\alpha$: $7\alpha^2 = 3\alpha + 2$ $\alpha^2 = \frac{3\alpha + 2}{7}$

Now, substitute this into the term $1-\alpha^2$: $1-\alpha^2 = 1 - \frac{3\alpha + 2}{7} = \frac{7 - (3\alpha + 2)}{7} = \frac{7 - 3\alpha - 2}{7} = \frac{5 - 3\alpha}{7}$

So, the term $\frac{\alpha}{1-\alpha^2}$ becomes: $\frac{\alpha}{(5-3\alpha)/7} = \frac{7\alpha}{5-3\alpha}$ Similarly, $\frac{\beta}{1-\beta^2} = \frac{7\beta}{5-3\beta}$.

Now, the expression $E$ is: $E = \frac{7\alpha}{5-3\alpha} + \frac{7\beta}{5-3\beta}$ Factor out 7: $E = 7 \left( \frac{\alpha}{5-3\alpha} + \frac{\beta}{5-3\beta} \right)$

Combine the terms inside the parenthesis: $E = 7 \left( \frac{\alpha(5-3\beta) + \beta(5-3\alpha)}{(5-3\alpha)(5-3\beta)} \right)$

Numerator inside parenthesis: $5\alpha - 3\alpha\beta + 5\beta - 3\alpha\beta = 5(\alpha+\beta) - 6\alpha\beta$

Substitute $\alpha+\beta = 3/7$ and $\alpha\beta = -2/7$: $5(3/7) - 6(-2/7) = 15/7 + 12/7 = 27/7$

Denominator inside parenthesis: $(5-3\alpha)(5-3\beta) = 25 - 15\beta - 15\alpha + 9\alpha\beta = 25 - 15(\alpha+\beta) + 9\alpha\beta$

Substitute $\alpha+\beta = 3/7$ and $\alpha\beta = -2/7$: $= 25 - 15(3/7) + 9(-2/7) = 25 - 45/7 - 18/7 = 25 - (45+18)/7 = 25 - 63/7 = 25 - 9 = 16$

Substitute these back into the expression for E: $E = 7 \left( \frac{27/7}{16} \right) = 7 \times \frac{27}{7 \times 16} = \frac{27}{16}$

Both methods yield the same result.