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In a quadratic equation kx^2-6x-1=0, determine …

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In a quadratic equation kx^2-6x-1=0, determine the values of k for which the equation does not have any real root.

Posted by Jiaur Choudhury 5 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Jay Darji 5 years, 11 months ago

K is less than -9

0Thank You

Yogita Ingle 5 years, 11 months ago

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

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

To find, the value of k = ?

{tex}∴ D=b^{2} -4ac{/tex}

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

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

=36 + 4k < 0

⇒ 4k < -36

⇒ k < {tex}\dfrac{-36}{4}{/tex}

⇒ k < - 9

∴ k < - 9

Hence, the value of k is < - 9.

0Thank You

Sarthak Bansal 5 years, 11 months ago

K is less than -9

0Thank You

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