Grade 11Algebra

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

Radhika Batra

12 Years agoGrade 11

12 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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