Question

find the zeros of the quadratic polynomial 6 x square - 7 x minus 3 and verify the relationship between the zeros and the coefficients​

Answer 1

$\large\underline{ \underline{ \sf \: Solution : \: \: \: }}$

We have ,

$\to$6x² - 7x - 3

By Prime factorisation method ,

$\sf \implies 6 {x}^{2} - 7x - 3 = 0 \\ \\ \sf \implies </p><p>6 {x}^{2} - 9x + 2x - 3 = 0 \\ \\ \sf \implies </p><p>3x(2x - 3) + 1(2x - 3) =0 \\ \\ \sf \implies </p><p>(3x+1)(2x-3)= 0 \\ \\ \sf \implies </p><p>x = - \frac{1}{3} \: \: or \: \: x = \frac{3}{2} </p><p>$

Therefore , the zeroes of 6x² - 7x - 3 are -1/3 and 3/2

Now ,

$\star \: \sf Sum \: of \: zeroes = \frac{7}{6} = - \frac{ (coefficient \: of \: x)}{coefficient \: of \: {x}^{2} } </p><p>$

$\star \: \sf Product \: of \: zeroes = - \frac{ \: 3 \: }{6} = \frac{ constant \: term }{coefficient \: of \: {x}^{2} }$

Hence verified