Quadratic Equations Question - None

Question

If $\alpha$ and $\beta$ are the roots of equation $2x^2 - 5x + 7 = 0$, then the equation whose roots are $2\alpha + 3\beta$ and $3\alpha + 2\beta$ is

Accepted Answer

Step 1: Since $\alpha$ and $\beta$ are roots of $2x^2 - 5x + 7 = 0$, we have:

$\alpha + \beta = \frac{5}{2}$ and $\alpha\beta = \frac{7}{2}$

Step 2: Find sum of new roots:

$(2\alpha + 3\beta) + (3\alpha + 2\beta) = 5(\alpha + \beta) = 5 \cdot \frac{5}{2} = \frac{25}{2}$

Step 3: Find product of new roots:

$(2\alpha + 3\beta)(3\alpha + 2\beta) = 6\alpha^2 + 4\alpha\beta + 9\alpha\beta + 6\beta^2$

$= 6(\alpha^2 + \beta^2) + 13\alpha\beta$

$= 6[(\alpha + \beta)^2 - 2\alpha\beta] + 13\alpha\beta$

$= 6\left[\left(\frac{25}{4}\right) - 7\right] + \frac{91}{2}$

$= 6 \cdot \frac{-3}{4} + \frac{91}{2} = \frac{82}{2} = 41$

Step 4: The equation with sum $\frac{25}{2}$ and product $41$ is:

$x^2 - \frac{25}{2}x + 41 = 0$ or $2x^2 - 25x + 82 = 0$

∴ Answer is (a).

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