If the roots of the equation $(b - c){x^2} + (c - a)x + (a - b) = 0$ be equal then b is equal to
- $\displaystyle {{a + c} \over 2}$
- $\displaystyle {{a - c} \over 2}$
- $\displaystyle{{a + c} \over {2ac}}$
- $\displaystyle{{a - c} \over {2ac}}$
For a quadratic equation Ax^2 + Bx + C = 0 to have equal roots, the discriminant B^2 - 4AC must be zero. Here, (c-a)^2 - 4(b-c)(a-b) = 0. Expanding this leads to (a+c-2b)^2 = 0, so a+c = 2b, or b = (a+c)/2.
For the quadratic equation (b minus c)x squared plus (c minus a)x plus (a minus b) equals 0, if the roots are equal, its discriminant must be zero. Setting the discriminant D equal to (c minus a) squared minus 4 multiplied by (b minus c) multiplied by (a minus b) equal to zero, we expand to get c squared minus 2ac plus a squared minus 4 times (ab minus b squared minus ac plus bc) equals 0. This simplifies to a squared plus 2ac plus c squared minus 4ab plus 4b squared plus 4bc minus 4ac, which rearranges to (a plus c) squared minus 4b multiplied by (a plus c) plus 4b squared equaling 0. Recognizing this as a perfect square, we get (a plus c minus 2b) squared equals 0, which means a plus c equals 2b, or b equals (a plus c) divided by 2.