- Home
- Standard 11
- Mathematics
4-2.Quadratic Equations and Inequations
medium
If the product of roots of the equation ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$ is $7$, then its roots will real when
A
$k = 1$
B
$k = 2$
C
$k = 3$
D
None of these
(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
