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If the product of x2 – 3kx + 2 e2lnk – 1 = 0 is 7, then the roots are real for k = . . . . . . . . . . . . . . . . . . . .

Radhika Batra , 10 Years ago
Grade 11
anser 1 Answers
Kevin Nash

Last Activity: 10 Years ago

The given equation is x2 – 3kx + 2e2lnk – 1 = 0
Or x2 – 3kx + (2k2 - 1 ) = 0
Here product of roots = 2k2 – 1
∴ 2k2 – 1 = 7 ⇒ k2 = 4 ⇒ k = 2, -2
Now for real roots we must have D ≥ 0
⇒ 9k2 – 4(2k2 – 1) ≥ 0 ⇒ k2 + 4≥ 0
Which is true for all k. Thus k = 2, -2
But for k = -2, in k is not defined
∴ Rejecting k = -2, we get k = 2.

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