Question

In a quadratic equation kx2-6x-1=0 determine the value of k which does not have real roots

Answer 1

in the quadratic equation Kx^2 - 6x - 1 = 0

a= K

b = -6

c = -1

since it does not have real roots

so

D < 0

b^2 - 4ac < 0

(-6)^2 - 4 × k(-1) < 0

36 + 4k < 0

36 + 4k - 36 < - 36

4k < -36

k < -9

so the value of k will be any number less than -9 .

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Answer 2

The value of k is < - 9.

Explanation:

The given quadratic equation is $kx^{2} -6x-1=0$

Here, a = k, b = - 6 and c = - d

To find,  the value of k = ?

∴ $D=b^{2} -4ac$

$=(-6)^{2} -4(k)(-1)$

$=36+4k$, does not have real roots.

=36 + 4k < 0

⇒ 4k < -36

⇒ k < $\dfrac{-36}{4}$

⇒ k < - 9

∴  k < - 9

Hence, the value of k is < - 9.