Question

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

Answer 1

Answer:

Step-by-step explanation:

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→$Given\:p(x)=3y^2+ky+12=0.$

⇒$a=3\\b=k\\c=12$

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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Answer 2

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