Question

if alpha and beta are zeros of polynomial 6 x square - 7 x minus 3, then form a quadratic polynomial whose zeros are 2 alpha and 2beta

Answer 1

$6 {x}^{2} - 7x - 3 \\ = 6 {x}^{2} - 9x + 2x - 3 \\ = 3x(2x - 3) + 1(2x - 3) \\ = (3x + 1)(2x - 3) \\ this \: implies \: x = \frac{ - 1}{3} \: or \: \frac{3}{2} \\ let \: \alpha = \frac{ - 1}{3} \: and \: \beta = \frac{3}{2} \\ then \: \\ 2 \times \alpha = \frac{ - 2}{3} \: and \: 2 \times \beta = 3 \\ now \: 2 \times \alpha + 2 \times \beta = \frac{7}{9} = sum\\ and \: 2 \times \alpha \times 2 \times \beta = \frac{ - 2}{3} \times 3 = - 2 = product \\ we \: know \\ a \: quadratic \: equation \: is \: of \: the \: form \: {x}^{2} - (sum ) \times x + (product) \\ therefore \: the \: equation \: is \\ {x}^{2} - ( \frac{7}{9} ) \times x + ( - 2) = {x}^{2} - \frac{7x}{9} - 2 \\ = 9 \times ( {x}^{2} - \frac{7x}{9} - 2) = 9 {x}^{2} - 7x - 18.$

Answer 2

Answer:

Step-by-step explanation: