If the product of the roots of the equation x 2 - 2 2 ? k x + 2 e 2 log k - 1 = 0 is 31, then the roots of the equation are real for k =

Question

If the product of the roots of the equation $x^{2} - 2 \sqrt{2} k x + 2 e^{2 \log k} - 1 = 0$ is 31, then the roots of the equation are real for $k =$

Smart Achievers · Accepted Answer

(3)

$Let α, β be the roots of$

$x^{2} - 2 \sqrt{2} k x + 2 e^{2 \log k} - 1 = 0,$

$α β = 2 e^{2 \log k} - 1$ ...(i)

$Product of roots (α β) is given 31$

$Hence from (i), 2 e^{2 \log k} - 1 = 31$ $or 2 e^{\log k^{2}} - 1 = 31$

$or 2 k^{2} = 32 or k^{2} = 16 or k = \pm 4$

$As \log (- 4) is not defined, so k = 4 .$

Solution

Q.
If the product of the roots of the equation $x^{2} - 2 \sqrt{2} k x + 2 e^{2 \log k} - 1 = 0$ is 31, then the roots of the equation are real for $k =$

1 - 4

2 1

3 4

4 0

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Solution

Q. If the product of the roots of the equation x2-22 kx+2e2logk-1=0 is 31, then the roots of the equation are real for k=

1 - 4

2 1

3 4

4 0

Ans. (3) Let α,β be the roots of x2-22 kx+2e2logk-1=0, αβ=2e2logk-1 ...(i) Product of roots (αβ) is given 31 Hence from (i), 2e2logk-1=31 or 2elogk2-1=31 or 2k2=32 or k2=16 or k=±4 As log(-4) is not defined, so k=4.

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Q.
If the product of the roots of the equation $x^{2} - 2 \sqrt{2} k x + 2 e^{2 \log k} - 1 = 0$ is 31, then the roots of the equation are real for $k =$