Question

find quadratic equation whose zeroes are 2/3 and -5/7​

Answer 1

0= x² - (a+b)x + (a*b)

0= x² - (2/3 -5/7)x - 10/21

0= x² -(-1/21)x - 10/21

0= 21x² + x -10

hence quadratic equation is

21x²+x-10=0

Thanks

Answer 2

HERE IS YOUR ANSWER

GIVEN:

$\alpha = \frac{2}{3 } \\ \beta = \frac{ - 5}{7} \\ therefore \\ \alpha + \beta = \frac{2}{3} - \frac{5}{7 } \\ \alpha + \beta = - \frac{1}{21 } \\ again \\ \alpha \beta = \frac{2}{3} \times \frac{ - 5}{ 7 } \\ \alpha \beta = \frac{ - 10}{21}$

$we \: know \: that \\ {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0 \\ {x}^{2} - ( - \frac{1}{21} )x - \frac{10}{21} = 0 \\ 21 {x}^{2} + x - 10 = 0$

HOPE IT HELPS YOU :)