Question

For what value of k, the equation 9x^2+6kx+4=0 has equal roots?​

Answer 1

Answer:

k = 2

Step-by-step explanation:

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Answer 2

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Given:

• Quadratic equation : $p(x) = 9x^2 + 6kx + 4 = 0$

To Find:

• Value of 'a' such that p(x) has equal roots

Concept:

We know that, if a quadratic polynomial has equal roots, then the value of Discriminant(D) is equal to 0.

Solution:

$\implies p(x) = 9x^2 + 6kx + 4 = 0 \\\\ D = 0 \\\\\implies B^2 - 4(A)(C) = 0 \\\\\text{Here, A = coefficient of x^2 , B = coefficient of x and C = constant.} \\\\\implies A = 9, B = 6k, C = 4 \\\\\implies (6k)^2 - 4(9)(4) = 0 \\\\\implies 36k^2 - 144 = 0 \\\\\implies 36k^2 = 144 \\\\\implies k^2 = \dfrac{144}{36} \\\\\implies k^2 = 4 \\\\\implies \bf{k = \pm 2}$

Conclusion:

Hence, for k = ±2 , p(x) has equal roots.

Learn More:

If, D > 0 => Real and distinct roots exist.

If, D = 0 => Real and equal roots exist.

If, D < 0 => No real roots exist. (imaginary roots exist)

Hope it helps u!!