Question

what is the value of k, if one root of quadratic equation kx^2-24x+28=0 is 6 +√22​

Answer 1

Answer:

Step-by-step explanation:

Hope it will help you.

Answer 2

Answer:

The value of k is  2.

Step-by-step explanation:

The given quadratic equation is $kx^2-24x+28=0$ and one root of the equation is $6+\sqrt{22}$.

We are required to find the value of k.

Substituting the value of given root in the quadratic equation,we get

$k(6+\sqrt{22})^2-24(6+\sqrt{22})+28=0\\= > k(36+22+12\sqrt{22})-144-24\sqrt{22}+28=0\\= > (58+12\sqrt{22})k=116+24\sqrt{22}\\= > k=\frac{116+24\sqrt{22}}{58+12\sqrt{22}} =2$

Therefore,

The value of k is found to be 2.

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