1.

Find the value of K kx^2×(6k+2)x+16=0​

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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