First slide

Theory of equations

Question

If the products of roots of the equation $x^{2} - 3 kx + 2 e^{2 \log k} - 1 = 0$
is 7, then its roots will real when

Easy

A

$k = 1$
B

$k = 2$
C

$k = 3$
D

None of these

Solution

Given equation can be written as
$x^{2} - 3 kx + 2 k^{2} - 1 = 0$
So the product of roots is $2 k^{2} - 1$ . But the product of roots is 7. Hence $2 k^{2} - 1 = 7 \Rightarrow 2 k^{2} = 8 \Rightarrow k = \pm 2$
But k cannot be negative.

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