Question

If one of the roots of equation ${{x}^{2}}+ax+3=0$ is 3 and one of the roots of the equation ${{x}^{2}}+ax+b=0$ is three times the other root, then the value of b is
A . 3
B. 4
C. 2
D. 1

Answer 1

Hint: In this question, we are given quadratic equations with their roots and we have to find the value of b. For this, first, we compare the equations with the standard form of the quadratic equation and then find the sum and the product of roots. By equating the roots we are able to get the value of b.

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

Complete step by step Solution:
Given equation is ${{x}^{2}}+ax+3=0$
Compare it with standard form of quadratic equation $a{{x}^{2}}+bx+c=0$, we get
$a=1,b=a,c=3$
Let $\alpha $ and 3 be the roots of the above equation
then the sum of roots ($\alpha +3$) = $-\dfrac{b}{a}$= $-\dfrac{(a)}{1}$= - a
and the product of roots ($3\alpha $) = $\dfrac{c}{a}$= $\dfrac{3}{1}$= 3
as $3\alpha $= 3
then $\alpha =1$
as $\alpha +3$= -a
By putting the value $\alpha =1$ in the above equation, we get
 a = - 4
Now let $\beta $ and $3\beta $ are the roots of the equation ${{x}^{2}}+ax+b=0$
Similarly we Compare it with standard form of quadratic equation $a{{x}^{2}}+bx+c=0$, we get
$a=1,b=a,c=b$
then the sum of roots ($\beta +3\beta $) = $-\dfrac{b}{a}$= $-\dfrac{(a)}{1}$= - a
$4\beta $= -a
As we find out the value of a =-4, so $\beta $= 1
and the product of roots ($\beta \times 3\beta $) = $\dfrac{c}{a}$= $\dfrac{b}{1}$= b
as $\beta $=1 , so $1\times 3=b$
then $3=b$
Hence the value of b = 3

Therefore, the correct option is (A).

Note: In these types of questions, we can find the sum and the product of roots by the formula
 if x and y are the roots of any quadratic equation the value of xy will be equal to $\dfrac{ constant\, term}{coefficient\, of\, x^2}$ and the sum of the roots that is x + y is equal to $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$ and by solving it we get the desired answer.