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