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If quadratic equation x²+k(2x+k-1)+2=0 has real and equal roots then find value of k. |
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Answer» ong>Answer: k=2 Step-by-step explanation: GIVEN, x2+k(2x+k−1)+2=0 Simplify above equation: x2+2kx+(k2−k+2)=0 Compare given equation with the general form of QUADRATIC equation, which ax2+bx+c=0 here, a=1,b=2k,c=(k2−k+2) Find discriminant: D=b2−4ac =(2k)2−4×1×(k2−k+2) =4k2−4k2+4k−8 =4k−8 Since roots are REAL and EQUAL (given) Put D=0 4k−8=0 k=2 |
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