First slide

Theory of equations

Question

Given that $α, γ$ are roots of the equation ${Ax}^{2} - 4 x + 1 = 0$ , and $β, δ$ the roots of the equation of ${Bx}^{2} - 6 x + 1 = 0$ , such that $α, β, γ, and δ$ are in H.P., then

Moderate

A

A = 3
B

A = 4
C

B = 2
D

B = 8

Solution

Since $α, β, γ, δ$ are in H.P., $1 / α, 1 / β, 1 / γ, 1 / δ$ are in A.P. and they maybe taken as $a - 3 d, a - d, a + d, a + 3 d$ . Replacing x by 1/x, we get the equation whose roots are $1 / α, 1 / β, 1 / γ, 1 / δ$ . Therefore, equation $x^{2} - 4 x + A = 0$ has roots a - 3d, a + d and equation $x^{2} - 6 x + B = 0$ has roots a - d, a + 3d. Sum of the roots is
$\begin{array}{ll} 2 (a - d) = 4, 2 (a + d) = 6 \\ ∴ a = 5 / 2, d = 1 / 2 \end{array}$
Product of the roots is
$\begin{aligned} (a - 3 d) (a + d) = A = 3 \\ (a - d) (a + 3 d) = B = 8 \end{aligned}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App