Math, asked by Ney13, 9 months ago

Find the value of k for which the given equation has real root:-
9x²+3kx+4=0.

Anonymous: Yeah I do, but I never learn lol

Ney13: Yeah,even I saw a few of them here.

Anonymous: I guess it's because I learned a lil Japanese before, Korean seems hard before it

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Answers

Answered by khyati4267

Given quadratic equation is= 9x² + 3kx + 4 = 0

On comparing with standard form of quadratic equation i.e ax² + bx + c =0,a≠0

Here, a = 9 , b= 3k, c= 4

D(discriminant)= b²-4ac

= (3k)² - 4× 9 ×4

= 9k² - 144

Since, roots of given equation are distinct. D > 0.

9k² - 144 > 0

9(k² - 16) >0

(k² - 16) >0 (9≠0)

k² -4²>0

(k-4) (k+4) >0

[ a² - b² = (a-b)(a+b)]

k > 4 and k< -4

Hence, the value of k is k > 4 and k< -4.

HOPE THIS WILL HELP YOU..

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