Question

If a and b are the roots of the equation ${{x}^{2}}-px+q=0$, then the value of $\dfrac{1}{a}+\dfrac{1}{b}$ is

Answer 1

Hint: In this question, we are given a quadratic equation with its roots and we have to find the value of $\dfrac{1}{a}+\dfrac{1}{b}$ . First, we compare the given equation with the standard form of the quadratic equation and then find the sum of roots and the product of roots of the given quadratic equation. Then we take the LCM of $\dfrac{1}{a}+\dfrac{1}{b}$and by putting the values we find the value and choose the correct option.

Formula used:
Sum of roots = $-\dfrac{b}{a}$
the product of roots = $\dfrac{c}{a}$

Complete step by step Solution:
Given quadratic equation is ${{x}^{2}}-px+q=0$…………………….. (1)
Compare it with standard form of quadratic equation $a{{x}^{2}}+bx+c=0$, we get
$a=1,b=-p,c=q$
Consider a and b are the roots of the equation (1)
then the sum of roots ($a+b$) = $-\dfrac{b}{a}$= $-\dfrac{(-p)}{1}$= p
and the product of roots $(ab)$ = $\dfrac{c}{a}$= $\dfrac{q}{1}$= q
Here we have to find the value of $\dfrac{1}{a}+\dfrac{1}{b}$

$\dfrac{1}{a}+\dfrac{1}{b}$= $\dfrac{a+b}{ab}$
Now we put the value of ($a+b$) and $(ab)$ in the above equation, we get
$\dfrac{a+b}{ab}$= $\dfrac{p}{q}$
Hence the value of $\dfrac{1}{a}+\dfrac{1}{b}$= $\dfrac{p}{q}$

Note: We can also use an alternate method to solve this question:-
As the question told ${{x}^{2}}-px+q=0$……………….. (1)
The above equation has roots a and b
Thus we can factorize the equation as
${{x}^{2}}-px+q=0$= $(x-a)(x-b)$
Which is equal to ${{x}^{2}}-(a+b)x+ab$………………… (2)
By comparing equation (1) with equation (2), we get
$p=a+b$ and $q=ab$
As we have to find the value of $\dfrac{1}{a}+\dfrac{1}{b}$
$\dfrac{1}{a}+\dfrac{1}{b}$= $\dfrac{a+b}{ab}$
Now we put the value of ($a+b$) and $(ab)$ in the above equation, we get
$\dfrac{a+b}{ab}$= $\dfrac{p}{q}$
You can use any method to solve this question.