Question

Find the values of $a$ and $b$ for which $x=\frac{3}{4}$ and $x=-2$ are roots of the equation $a{x}^2+bx-6=0$

Answer 1

Put $x=\frac{3}{4}$ in $a{x}^2+bx-6=0$

$\Rightarrow a(\frac{9}{16})+b(\frac{3}{4})=6$

$\Rightarrow 9a+12b=96$ ......(i)

Put $x=-2$

$\Rightarrow a{x}^2+bx-6=0$

$\Rightarrow 4a-2b=6$ ......(ii)

Solve eqn(i) & (ii) by multiplying eqn (ii) by 6 and then subtracting it from eqn(i)

We get, $a=4$

Put in eqn.(ii), $16-2b=6$

$16-6=2b$

$10=2b$

$b=5$