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Find the value of k for which the equation (3k+1)x^2+(k+1)x+k=0 has real and equal roots |
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Answer» Answer: Step-by-step EXPLANATION: so d=0 b²= 4ac (k+1)²= 4*( 3k+1)*k k²+1 +2k = 4*( 3k²+k) 11k²+2k-1 =0 k= (-2±√48)÷22 |
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