I need ans for this in detailed form plzz
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Given Equation is (3k + 1)x^2 + 2(k + 1)x + k = 0.
On comparing with ax^2 + bx + c = 0, we get
a = (3k + 1), b = 2(k + 1), c = k.
Now,
Given that The Equation has real roots.
b^2 - 4ac = 0
(2(k + 1))^2 - 4(3k + 1)(k) = 0
4(k^2 + 1 + 2k) - 4(3k^2 + k) = 0
4k^2 + 4 + 8k - 12k^2 - 4k = 0
-8k^2 + 4k + 4 = 0
-2k^2 + k + 1 = 0
-2k^2 - 1k + 2k + 1 = 0
-k(2k + 1) + 1(2k + 1) = 0
(-k + 1)(2k + 1) = 0
(-k + 1) = 0 (or) (2k + 1) = 0
k = 1 (or) k = -1/2.
Hope this helps!
On comparing with ax^2 + bx + c = 0, we get
a = (3k + 1), b = 2(k + 1), c = k.
Now,
Given that The Equation has real roots.
b^2 - 4ac = 0
(2(k + 1))^2 - 4(3k + 1)(k) = 0
4(k^2 + 1 + 2k) - 4(3k^2 + k) = 0
4k^2 + 4 + 8k - 12k^2 - 4k = 0
-8k^2 + 4k + 4 = 0
-2k^2 + k + 1 = 0
-2k^2 - 1k + 2k + 1 = 0
-k(2k + 1) + 1(2k + 1) = 0
(-k + 1)(2k + 1) = 0
(-k + 1) = 0 (or) (2k + 1) = 0
k = 1 (or) k = -1/2.
Hope this helps!
siddhartharao77:
:-)
tq
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