Question

If the zeroes of the rational expression $(ax+b)(3x+2)$ are $-\frac{2}{3}$ and $\frac{1}{2}$, then $a+b=$

• A. $4$
• B. $0$
• C. $-b$
• D. None of these

Answer 1

The correct option is B $0$

$(ax+b)(3x+2)=0$

$\Rightarrow$ $3a{x}^2+2ax+3bx+2b$

$\Rightarrow$ $3a{x}^2+(2a+3b)x+2b=0$

It is given that $\frac{-2}{3}$ and $\frac{1}{2}$ are zeros of the equation $3a{x}^2+(2a+3b)x+2b=0$

Here, $A=3a,B=2a+3b,C=2b$

Let $\alpha =\frac{-2}{3}$ and $\beta =\frac{1}{2}$

We know,

$\Rightarrow$ $\alpha +\beta =\frac{-B}{A}$

$\Rightarrow$ $\frac{-2}{3}+\frac{1}{2}=\frac{-(2a+3b)}{3a}$

$\Rightarrow$ $\frac{-4+3}{6}\times 3a=-2a-3b$

$\Rightarrow$ $\frac{-a}{2}=-2a-3b$

$\Rightarrow$ $-a=-4a-6b$

$\Rightarrow$ $3a+6b=0$ ----- ( 1 )

Now,

$\Rightarrow$ $\alpha .\beta =\frac{CA}{A}$

$\Rightarrow$ $\frac{-2}{3}\times \frac{1}{2}=\frac{2b}{3a}$

$\Rightarrow$ $\frac{-1}{3}\times 3a=2b$

$\Rightarrow$ $-a+2b=0$ ----- ( 2 )

Adding equation ( 1 ) and ( 2 ) we get,

$b=0$

Put $b=0$ in ( 2 ) we get,

$a=0$

$\therefore$ $a+b=0+0=0$