Math, asked by dalvirekha3, 6 months ago

The roots of quadratic equation 3y2+ky + 12= 0,are real and equal. Complete Avticty to find value of k​

Answers

Answered by Itzraisingstar
44

Answer:

Step-by-step explanation:

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Well I can help u with this answer:

Given\:p(x)=3y^2+ky+12=0.

a=3\\b=k\\c=12

It is given that roots of p(x) are real and equal

So:

b^2-4ac=0

(k)^2-4(3)(12)=0

k^2=144,\\\\

⇒k=12.

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Answered by anthonypaulvilly
0

Answer:

         a) 12

Step-by-step explanation:

3y² + ky + 12 = 0

if it has equal roots, then the determinant = 0

determinant of a quadratic equation = √(b² - 4ac) = 0

a = 3 , b = k , c = 12

√(b² - 4ac) = 0

√(k² - 4 × 3 × 12) = 0

√(k² - 144) = 0   {square both sides}

k² - 144 = 0

k² = 144

k = 12

⇒ a) 12

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