Find k for real solutions of quadratic equation

Blaine Jimenez

Blaine Jimenez

Answered question

2022-04-13

Find k for real solutions of quadratic equation sin4x+2(1+k)sin2xk2=0

Answer & Explanation

Latkaxu8j

Latkaxu8j

Beginner2022-04-14Added 11 answers

Step 1
Note that since 0 and 1 enclose both the roots, and the given quadratic expression has positive leading coefficient, so the quadratic expression should take non-negative sign at both 0 and 1, or at least one of them should be the root. So if f(t) denotes this quadratic expression, we have f(0)0, and f(1)0. Hence
(k2)0 and k+10
which is impossible, because
k+10k+21k21
Hence no such k exists.
xmzdenisemst0

xmzdenisemst0

Beginner2022-04-15Added 6 answers

Step 1
Express the equation as
k=22sin2xsin4x2sin2x1
=14cos2x3cos2x6
For x in the real domain, k has the range k(-2,-1)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?