Math, asked by lingadurai22062006, 5 months ago

Find the value of K kx^2×(6k+2)x+16=0
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Answered by Anonymous
9

Answer:

So, the values of k are 2 or 3. (ii) kx2 +(6k + 2)x + 16 = 0. a = k, b = 6k + 2 and c = 16. If the roots are real and equal, then Δ = 0. Δ = b2 - 4ac. (6k + 2)2 - 4(k)(16) = 0. (6k)2 + 2(6k)(2) ............

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