First slide

Theory of equations

Question

$If the roots of equation x^{2} + q x + p = 0 are each 1 less than the roots of the equation x^{2} + p x + q = 0, then (p + q) is equal to$

Moderate

A

-2
B

-4
C

-5
D

-6

Solution

$Let the roots of the equation x^{2} + q x + p = 0 be α - 1 and β - 1 .$
$α + β - 2 = - q - - - - (1)$
$(α - 1) (β - 1) = p - - - - - - (2)$
$Then roots of the equation x^{2} + p x + q = 0 are α and β .$
$α + β = - p - - - - - (3)$
$α β = q - - - - - (4)$
$\begin{aligned} From (1) and (3), p - q = - 2 \\ From (2), α β - (α + β) + 1 = p \end{aligned}$
$\begin{array}{l} ∴ q + p + 1 = p \\ or q = - 1 \end{array}$
$On solving, we get p = - 3 .$

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