First slide

Theory of equations

Question

$Consider the equation (x^{2} - 2 x + m) (x^{2} - 2 x + n) = 0 (where m and n are real numbers). Let the roots α, β, γ, δ (α < β < γ < δ)$
$of the equation form an A.P. with first term equal$ $to \frac{1}{4} . Then the value of \frac{m}{n} is$

Moderate

A

$\frac{7}{15}$
B

$\frac{8}{15}$
C

3
D

$\frac{1}{2}$

Solution

$Given first term being A = \frac{1}{4}$
$So, 4 roots in A.P. are \frac{1}{4}, \frac{1}{4} + d, \frac{1}{4} + 2 d, \frac{1}{4} + 3 d$
$Now given, (x^{2} - 2 x + m) (x^{2} - 2 x + n) = 0$
We can see that the sum of the roots in both the factors is 2.
$\begin{aligned} Hence, \frac{1}{4} + \frac{1}{4} + d + \frac{1}{4} + 2 d + \frac{1}{4} + 3 d = 4 \\ ∴ d = \frac{1}{2} \end{aligned}$
$\begin{array}{l} Hence, roots are \frac{1}{4}, \frac{3}{4}, \frac{5}{4} and \frac{7}{4} . \\ Here, \frac{3}{4} + \frac{5}{4} = 2 and \frac{1}{4} + \frac{7}{4} = 2 \\ Hence, n = \frac{3}{4} \times \frac{5}{4} = \frac{15}{16} \\ and m = \frac{1}{4} \times \frac{7}{4} = \frac{7}{16} \end{array}$

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