Question

if alpha and beta are the zeros of polynomial x^2+4x-3 form a quadratic equation whose zeroes are (3+alpha) and (3+beta).

Answer 1

${x}^{2} + 4x - 3 \\ by \: spliting \: the \: middle \: term \\ {x}^{2} + 3x + x - 3 \\ x(x + 3) + 1(x - 3) \\ (x - 3) \: (x + 1) \\ x = 3 \: \: x = - 1 \\ \alpha = 3 \\ \beta = - 1 \\ for \: new \: quadratic \: eq. \\ (3 + \alpha ) \: \: \: \: \: (3 + \beta ) \\ (3 + 3) \: \: \: \: \: (3 + ( - 1)) \\ = 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 \\ \alpha = 6 \\ \beta = 2 \\ by \: quadatic \: formula \\ {x}^{2} + ( \alpha + \beta ) - \alpha . \beta \\ {x}^{2} + (6 + 2) -( 6)(2 )\\ {x}^{2} + 8x - 12 \\ this \: is \: your \: answer \\ hope \: it \: will \: help \: u$