First slide

Theory of equations

Question

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

Moderate

A

3,8
B

4,7
C

2,7
D

None of these

Solution

$α, β, γ, δ are in H.P. \Rightarrow \frac{1}{α}, \frac{1}{β}, \frac{1}{γ}, \frac{1}{δ} are in A.P. and they may be taken as a - 3 d, a - d, a + d, a + 3 d .$
$Replacing x by 1 / x we get the equation whose roots are$ $\frac{1}{α}, \frac{1}{β}, \frac{1}{γ}, \frac{1}{δ} etc.$
$Thus, x^{2} - 4 x + A = 0 has roots a - 3 d, a + d .$
$And x^{2} - 6 x + B = 0 has roots a - d, a + 3 d .$
$\begin{array}{l} Sum = 2 (a - d) = 4, 2 (a + d) = 6 \\ ∴ a = 5 / 2, d = 1 / 2 \\ Product = (a - 3 d) (a + d) = A = 3 \\ (a - d) (a + 3 d) = B = 8 \end{array}$

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