Question

If a and b are zeroes of p(x)=6x^2+x-1.Find the value of 2(1/a+1/b)+3ab ​

Answer 1

Answer:

3/2

Step-by-step explanation:

Given,

a,b are the zeroes of p(x) = $6x^2+x-1$

To Find :-

Value of :-

$2\bigg(\dfrac{1}{a}+\dfrac{1}{b}\bigg)+3ab$

Solution :-

$\sf 6x^{2}+x-1$

According to question roots of the equation = a , b

We know that :-

$Sum\: of \:the \:roots = \dfrac{-(coefficient\:of\:'x'\:term)}{coefficient\:of\:x^2\:term}$

$product\:of\:the\:roots = \dfrac{constant\:term}{coefficient\:of\:x^2\:term}$

$\implies a+b = \dfrac{-1}{6}$

$a\times b=\dfrac{-1}{6}$

We need to find value of :-

$2\bigg(\dfrac{1}{a}+\dfrac{1}{b}\bigg)+3ab$

$2\bigg(\dfrac{b+a}{ab}\bigg)+3ab$

$2\bigg(\dfrac{a+b}{ab}\bigg)+3ab$

$2\left(\dfrac{\dfrac{-1}{6}}{\dfrac{-1}{6}}\right)+3\left(\dfrac{-1}{6}\right)$

$2(1)+\dfrac{-3}{6}$

$2-\dfrac{1}{2}$

$\dfrac{4-1}{2}$

$\dfrac{3}{2}$