A) 2
B) 0
C) 1
D) 3
Correct Answer: A
Solution :
Given equation \[(1+2k){{x}^{2}}+(1-2k)x+(1-2k)=0\] If equation is a perfect square then root are equal i.e., \[{{(1-2k)}^{2}}-4(1+2k)\,(1-2k)\,=0\] i.e., \[k=\frac{1}{2},\,\frac{-3}{10}\]. Hence total number of values = 2.You need to login to perform this action.
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