Question

If (2x - 1) is a factor for both 6x^2+ax-4 and bx^2 - 11x + 3 then the value of 'a' and 'b' are

Answer 1

$(2x - 1) \: is \: factor \: of \: 6 {x}^{2} + ax - 4 \: and \: b {x}^{2} - 11x + 3 \\ then \:by \: substituting \: x = \frac{1}{2} \: in \: these \: equations \\ 6 \times \frac{1}{2} \times \frac{1}{2} + a \times \frac{1}{2} - 4 = 0 \\ \frac{a}{2} = 4 - \frac{3}{2} = \frac{8 - 3}{2} = \frac{5}{2} \\ a = 5 \\ b \times \frac{1}{2} \times \frac{1}{2} - 11 \times \frac{1}{2} + 3 = 0 \\ \frac{b}{4} = \frac{11}{2} - 3 = \frac{11 - 6}{2} = \frac{5}{2} \\ b = \frac{5 \times 4}{2} = 10 \\ a = 5 \: \: \: \: \: \: \: \: \: \: \: \: \\ b = 10$