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Find roots of the equation by quadratic formula : x^(2)+x-(a+2)(a+1)=0 |
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Answer» Solution :The given equation is `x^(2)+x-(a+2)(a+1)=0` Comparing it with `AX^(2)+Bx+C=0`, we get `A=1,B=1andC=-(a+2)(a+1)` `:.x-B+-sqrt(B^(2)-4AC)/(2A)` `x=(-1+-sqrt(1^(2)-4xx1xx[-(a+2)(a+1)]))/(2XX1)` `impliesx=(-1+-sqrt(1+4(a^(2)+33a+2)))/(2)` `impliesx=(-1+-sqrt(1+4a^(2)+12a+8))/(2)` `impliesx=(-1+-sqrt(4a^(2)+12a+9))/(2)` `impliesx=(-1+-sqrt((2a+3)^(2)))/(2)` `impliesx=(-1+-(2a+3)^(2))/(2)` `impliesx=(-1+2a+3^(2))/(2)and (-1-2a-3)/(2)` `impliesx=(2a+2)/(2)and (-2a-4)/(2)` `impliesx=(a+1)and-(a+2)` are ROOTS of the equation. |
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