Question

Alpha-beta=4and alpha cube-beta cube=208, find the quadratic equation whose roots are alpha and beta

Answer 1

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Answer 2

$heya \\ \\ required \: quadratic \: equation \: in \: x \: is \\ \\ x {}^{2} - ( \alpha + \beta )x + ( \alpha \beta ) = 0 \\ \\ \alpha {}^{3} - \beta {}^{3} = ( \alpha - \beta ) {}^{3} + 3 \alpha \beta ( \alpha - \beta ) \\ \\ 208 = (4) {}^{3} + 3 \alpha \beta (4) \\ \\ \alpha \beta = 12 \\ \\ \\ \alpha - \beta = \sqrt{( \alpha + \beta ) {}^{2} - 4 \alpha \beta } \\ \\ 16 + 48 = ( \alpha + \beta ) {}^{2} \\ \\ \alpha + \beta = \sqrt{60} \\ \\ \\ so \: quadratic \: equation \: is \: \\ \\ x {}^{2} - (2 \sqrt{15} )x + 12 = 0$