Question

find the values of k for which the roots are real and equal in each of the following equations (3k+1)x^2+2(k+1)x+k=0
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Answer 1

hope it will help you:-)

Answer 2

Answer:

Here a= (3k+1), b= 2(k+1), c= k.

therefore, b^2-4ac= [2(k+1)]^2 - 4(3k+1)(k)

• =(2k+2)^2 - 4(3k^2 + k)
• =(2k)^2 + 2(2k)(2) + (2)^2 - 12k^2 + 4k
• =4k^2 + 8k + 4 - 12k^2 + 4k
• =4k^2 - 12k^2 + 8k + 4k + 4
• =-8k^2 + 12k + 4

Now, the roots are real and equal,

therefore, b^2-4ac = 0

• -8k^2 + 12k + 4 = 0
• -4(2k^2 - 3k - 1) = 0
• 2k^2 - 3k - 1 = 0÷-4
• 2k^2 - 3k - 1 = 0

Now, you have got a quadratic equation, solve it, you will get 2 value of k.