Question

If the roots of the equation $a{{x}^{2}}+bx+c=0$ are l and 2l, then
( a ) ${{b}^{2}}=9ac$
( b ) $2{{b}^{2}}=9ac$
( c ) ${{b}^{2}}=-4ac$
( d ) none of these

Answer 1

Hint:We are given an equation with its roots . we used the identity of sum of roots $-\dfrac{b}{a}$and the product of roots $\dfrac{c}{a}$. After solving these equations, we are able to get the desired equation.

Formula used:
We used the formula of the sum of its roots and the product of its roots
Sum of its roots = $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$
Product of its roots =$\dfrac{ constant\, term}{coefficient\, of\, x^2}$

Complete step by step Solution:
Given equation is $a{{x}^{2}}+bx+c=0$
L and 2l are its roots
We know that for any quadratic equation $a{{x}^{2}}+bx+c=0$
Sum of its roots = $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$=$\dfrac{ - b}{a}$
So, we get
L + 2l = $-\dfrac{b}{a}$
Solving further, we get 3l = $-\dfrac{b}{a}$
We get l = $2{{b}^{2}}=9ac$$-\dfrac{b}{3a}$
Squaring both sides, ${{l}^{2}}=\dfrac{{{b}^{2}}}{9{{a}^{2}}}$……………………………………………….. (1)
Also product of its roots $\dfrac{ constant\, term}{coefficient\, of\, x^2}$
So $l\times 2l=\dfrac{c}{a}$
That is $2{{l}^{2}}=\dfrac{c}{a}$
Now cross multiplying the above equation, we get
Then ${{l}^{2}}=\dfrac{c}{2a}$……………………………………………………….. (2)
Now we equate the equation (1) and (2), we get
$\dfrac{{{b}^{2}}}{9{{a}^{2}}}=\dfrac{c}{2a}$
Now cross- multiplying the above equation and solving it , we get
 we get $2{{b}^{2}}=9ac$

Therefore, the correct option is (b).

Note:Whenever we solve these types of questions where roots are given, we use the identity of the product of roots which is 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 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.