Question

Let λ=0 be a real number. Let α,β be the roots of the equation 14x^2−31x+3λ=0 and α,γ be the roots of the equation 35x^2−53x+4λ=0. Then 3α β and 4α γ are the roots of the equation

• A. 7x2−245x+250=0
• B. 49x2−245x+250=0
• C. 49x2+245x+250=0
• D. 7x2+245x−250=0

Answer 1

Answer: (B)

49x2−245x+250=0

Answer 2

Let the given equations be:

1. $14x^2 - 31x + 3λ = 0$ with roots $\alpha$ and $\beta$.
2. $35x^2 - 53x + 4λ = 0$ with roots $\alpha$ and $\gamma$.

Since $\alpha$ is a common root, it satisfies both equations:

$14\alpha^2 - 31\alpha + 3\lambda = 0$ (3)

$35\alpha^2 - 53\alpha + 4\lambda = 0$ (4)

Eliminate $\lambda$ from (3) and (4). Multiply (3) by 4 and (4) by 3:

$56\alpha^2 - 124\alpha + 12\lambda = 0$

$105\alpha^2 - 159\alpha + 12\lambda = 0$

Subtracting the first from the second:

$(105\alpha^2 - 159\alpha + 12\lambda) - (56\alpha^2 - 124\alpha + 12\lambda) = 0$

$49\alpha^2 - 35\alpha = 0$

$7\alpha(7\alpha - 5) = 0$

From equation (1), the product of roots is $\alpha\beta = \frac{3\lambda}{14}$.

From equation (2), the product of roots is $\alpha\gamma = \frac{4\lambda}{35}$.

If $\alpha = 0$, then $3\lambda/14 = 0 \implies \lambda = 0$ and $4\lambda/35 = 0 \implies \lambda = 0$. The question states $\lambda \neq 0$ (implied by the similar question context). Thus $\alpha \neq 0$.

So, $7\alpha - 5 = 0 \implies \alpha = \frac{5}{7}$.

Substitute $\alpha = \frac{5}{7}$ into equation (3):

$14(\frac{5}{7})^2 - 31(\frac{5}{7}) + 3\lambda = 0$

$14(\frac{25}{49}) - \frac{155}{7} + 3\lambda = 0$

$\frac{2 \times 25}{7} - \frac{155}{7} + 3\lambda = 0$

$\frac{50}{7} - \frac{155}{7} + 3\lambda = 0$

$-\frac{105}{7} + 3\lambda = 0$

$-15 + 3\lambda = 0 \implies 3\lambda = 15 \implies \lambda = 5$.

Now find the values of $\beta$ and $\gamma$.

From equation (1), $\alpha\beta = \frac{3\lambda}{14}$. Substitute $\alpha = \frac{5}{7}$ and $\lambda = 5$:

$\frac{5}{7}\beta = \frac{3 \times 5}{14} = \frac{15}{14}$

$\beta = \frac{15}{14} \times \frac{7}{5} = \frac{3}{2}$.

From equation (2), $\alpha\gamma = \frac{4\lambda}{35}$. Substitute $\alpha = \frac{5}{7}$ and $\lambda = 5$:

$\frac{5}{7}\gamma = \frac{4 \times 5}{35} = \frac{20}{35} = \frac{4}{7}$

$\gamma = \frac{4}{7} \times \frac{7}{5} = \frac{4}{5}$.

The question asks for the equation whose roots are $\frac{3\alpha}{\beta}$ and $\frac{4\alpha}{\gamma}$.

Let the new roots be $r_1 = \frac{3\alpha}{\beta}$ and $r_2 = \frac{4\alpha}{\gamma}$.

$r_1 = \frac{3 \times (5/7)}{3/2} = \frac{15/7}{3/2} = \frac{15}{7} \times \frac{2}{3} = \frac{5}{7} \times 2 = \frac{10}{7}$.

$r_2 = \frac{4 \times (5/7)}{4/5} = \frac{20/7}{4/5} = \frac{20}{7} \times \frac{5}{4} = \frac{5}{7} \times 5 = \frac{25}{7}$.

The sum of the new roots is $S = r_1 + r_2 = \frac{10}{7} + \frac{25}{7} = \frac{35}{7} = 5$.

The product of the new roots is $P = r_1 \times r_2 = \frac{10}{7} \times \frac{25}{7} = \frac{250}{49}$.

The quadratic equation with roots $r_1$ and $r_2$ is $x^2 - Sx + P = 0$.

$x^2 - 5x + \frac{250}{49} = 0$.

To obtain integer coefficients, multiply the equation by 49:

$49x^2 - 49 \times 5x + 49 \times \frac{250}{49} = 0$

$49x^2 - 245x + 250 = 0$.