Question

if zeros of rational expression (ax + b) (3x + 2) are -2/3 and 1/2 then a + b is​

Answer 1

$\large\underline{\bold{Given- }}$

$\sf \: A \: rational \: expresion \: (ax + b)(3x + 2) \: having \: zeroes \: - \dfrac{2}{3} \: and \: \dfrac{1}{2}$

$\large\underline{\bold{To\:Find - }}$

$\sf \: a + b$

$\large\underline{\bold{Solution-}}$

Concept Used :-

If α and β are zeroes of ax² + bx + c then

$\boxed{{\sf Sum\ of\ the\ zeroes= \alpha + \beta = \frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

OR

$\boxed{{\tt Sum\ of\ the\ zeroes= \alpha + \beta = \frac{-b}{a}}}$

And

$\boxed{{\sf Product\ of\ the\ zeroes= \alpha \beta = \frac{Constant}{coefficient\ of\ x^{2}}}}$

OR

$\boxed{{\tt Product\ of\ the\ zeroes= \alpha \beta = \frac{c}{a}}}$

Let's solve the problem now!!

$\sf \: Let \: f(x) \: = \: (ax + b)(3x + 2)$

$\sf \: = 3a {x}^{2} + 2ax + 3bx + 2b$

$\sf \: = 3a {x}^{2} + x(2a + 3b) + 2b$

So,

$\sf \:f(x) \: is \: a \: quadratic \: polynomial \: having \: zeros \: \dfrac{ - 2}{3} \: and \: \dfrac{1}{2}$

Hence,

• using product of zeroes, we have

${{\sf Product\ of\ the\ zeroes= \alpha \beta = \dfrac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\: - \dfrac{2}{3} \times \dfrac{1}{2} = \dfrac{2b}{3a}$

$\rm :\longmapsto\:\dfrac{2b}{a} = - 1$

$\rm :\implies\: \boxed{ \bf{a \: = \: - \: 2b}} - - (1)$

Again,

• Using sum of zeroes, we have

${{\sf Sum\ of\ the\ zeroes= \alpha + \beta = \dfrac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\sf \: - \dfrac{2}{3} + \dfrac{1}{2} = - \dfrac{(3b + 2a)}{3a}$

$\sf \: \dfrac{ - 4 + 3}{6} = - \dfrac{3b + 2( - 2b)}{3( - 2b)} \: \: \: \{ \because \: a \: = \: - \: 2b \}$

$\sf \: - \dfrac{1}{6} = - \dfrac{3b - 4b}{ - 6b}$

$\sf \: - \dfrac{1}{6} = - \dfrac{1}{6}$

Hence,

$\sf \: a + b = - 2b + b = - b$

$\bf \: Hence \: value \: of \: a \: + \: b \: = \: - \: b$