Question

Find the zeroes of x square-3x - 10=0
and verify
the relationship between its zeroes and
cofficient .​

Answer 1

$\huge\mathcal {\color {pink}{Answer}}$

${x}^{2} - 3x - 10 = 0$

$d = {b}^{2} - 4ac$

$d = {( - 3)}^{2} - 4(1)( - 10)$

$d = 9 + 40 = 49$

$x = \frac{ - b + - \sqrt{d} }{2a}$

$x = \frac{ - ( - 3) + - \sqrt{49} }{2(1)}$

$x = \frac{3 + - 7}{2}$

$( + ) \: \: \alpha = \frac{3 + 7}{2}$

$\alpha = \frac{10}{2}$

$\alpha = 5$

$( - ) \: \: \beta = \frac{ - ( - 3) - 7}{2(1)}$

$\beta = \frac{3 - 7}{2}$

$\beta = \frac{ - 4}{2}$

$\beta = - 2$

•°• zeroes are 5 & -2