Question

how to do Q-5 ,ch quadratic equ , class 10

Answer 1

Mark me brainlist. Hope it helps..
$a( \frac{2}{3})^{2} + 7( \frac{2}{3} ) + b = 0 \\ \frac{4a}{9} + \frac{14}{3} + b = 0......eq.1 \\ \\ a ({ - 3})^{2} + 7( - 3) + b = 0 \\ 9a - 21 + b = 0........eq.2 \\- (\frac{4a}{9} - \frac{14}{3} + b = 0) ...... eq. 1 \\ - - - - - - - - - - - - - \\ (9a - \frac{4a}{9} ) - 21 + \frac{14}{3} + ( b - b) = 0 \\ \frac{(81 - 4)a}{9} \times ( \frac{ - 63 + 14}{3} ) = 0 \\ \frac{77a}{9} - \frac{49}{3} = 0 \\ \frac{11a}{3} = 7 \\ a = \frac{21}{11} \\ \\ \\ 9a - 21 + b = 0 \\ 9 (\frac{21}{11}) - 21 + b = 0 \\ = > b = 21 - \frac{9 \times 21}{11} \\ = \frac{11 \times 21 - 9 \times 21}{11} = \frac{2 \times 21}{11} = \frac{42}{11} = b$