4-2.Quadratic Equations and Inequations
medium

જો સમીકરણ ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$ ના બીજનો ગુણાકાર $7$ હોય તો તેમના બીજ વાસ્તવિક છે કે જયાં 

A

$k = 1$

B

$k = 2$

C

$k = 3$

D

એકપણ નહી.

(IIT-1984)

Solution

(b) Given equation can be written as

${x^2} – 3kx + 2{k^2} – 1 = 0$

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.

Standard 11
Mathematics

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