Question

the root of the equation 9x²+7x-2=0​

Answer 1

Answer:

$9x ^{2} + 7x - 2 = 0 \\ \\ = > 9x ^{2} + (9 - 2)x - 2 = 0 \\ \\ = > 9x ^{2} + 9x - 2x - 2 = 0 \\ \\ = > 9x(x + 1) - 2(x + 1) = 0 \\ \\ = > (9x - 2)(x + 1) = 0 \\ \\ either \: 9x - 2 = 0 \\ \\ = > 9x = 2 \\ \\ = > x = \frac{2}{9} \\ \\ or \: x + 1 = 0 \\ \\ = > x = - 1$

▶ So option ( a ) = 2 / 9 and - 1 iS ✅

hope it helps uH ✨

Answer 2

${9x}^{2} + 7x - 2 = 0$

On comparing with standard equation,

A = 9, B = 7, C = - 2

$∴ \: \: \: \: \: \: From \: x = \frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a}$

$\: \: \: \: \: \: \: \: \: \: \: x = \frac{ - 7± \sqrt{ {7}^{2} - 4.9( - 2)} }{2.9}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 7± \sqrt{49 + 72} }{18}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 7± \sqrt{121} }{18}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - 7± 11}{18}$

On taking ( +) sign,

$\: \: \: \: \: \: \: \: \: \: \: \: x = \frac{ - 7 + 11}{18} = \frac{4}{18} = \frac{2}{9}$

On taking ( -) sign,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{ - 7 - 11}{18} = \frac{ - 18}{18} = - 1$

$∴ \: \: \: \: \: \: \: \: \: \: x = \frac{2}{9} , - 1.$