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# If the root of the equation $a{{x}^{2}}+bx+c=0$ are in the ratio m:n, then (a) $mn{{a}^{2}}=\left( m+n \right){{c}^{2}}$(b) $mn{{b}^{2}}=\left( m+n \right)ac$(c) $mn{{b}^{2}}=\left( m+n \right)2ac$(d) None of these

Last updated date: 20th Mar 2023
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Hint: Here we are given that the roots are in the ratio m:n, so consider the roots as $m\alpha$ and $n\alpha$. Now, use the sum of the roots $=\dfrac{-b}{a}$. From this, find the value of $\alpha$. Then substitute the value of $m\alpha$ in the given equation to get the desired result.

Complete step by step solution:

We are given that the roots of the equation $a{{x}^{2}}+bx+c=0$ are in the ratio m:n. We have to find the relation between m, n, a, b and c.
Let us first consider the quadratic equation given in the question.
$a{{x}^{2}}+bx+c=0....\left( i \right)$
We are given that the roots of the above equation are in ratio m:n. So, let us take the roots of the above equation to be $m\alpha$ and $n\alpha$.
We know that for any general quadratic equation, $a{{x}^{2}}+bx+c=0$, the sum of the root $=\dfrac{-\left( \text{coefficient of x} \right)}{\left( \text{coefficient of }{{\text{x}}^{2}} \right)}=\dfrac{-b}{a}$. So, we get,
$\left( m\alpha +n\alpha \right)=\dfrac{-b}{a}$
By taking $\alpha$ common from LHS of the above equation, we get,
$\alpha \left( m+n \right)=\dfrac{-b}{a}$
By dividing both sides of the above equation by (m + n) we get,
$\alpha =\dfrac{-b}{a\left( m+n \right)}....\left( ii \right)$
We know that $'m\alpha '$ is the root of equation (i). So let us substitute $x=m\alpha$ in the equation (i). We get,
$a{{\left( m\alpha \right)}^{2}}+b\left( m\alpha \right)+c=0$
We know that ${{\left( ab \right)}^{n}}={{a}^{n}}.{{b}^{n}}$. By applying this in the above equation, we get,
$a{{\left( m \right)}^{2}}.{{\left( \alpha \right)}^{2}}+b.m.\alpha +c=0$
By substituting the value of $\alpha$ from equation (ii) in the above equation, we get,
$a{{m}^{2}}{{\left( \dfrac{-b}{a\left( m+n \right)} \right)}^{2}}+b.m.\left( \dfrac{-b}{a\left( m+n \right)} \right)+c=0$
$\dfrac{a{{m}^{2}}{{b}^{2}}}{{{a}^{2}}{{\left( m+n \right)}^{2}}}-\dfrac{{{b}^{2}}m}{a\left( m+n \right)}+c=0$
By multiplying ${{a}^{2}}{{\left( m+n \right)}^{2}}$ on both sides of the above equation, we get,
$\Rightarrow a{{m}^{2}}{{b}^{2}}-{{b}^{2}}.m\left( a\left( m+n \right) \right)+{{a}^{2}}{{\left( m+n \right)}^{2}}c=0$
By simplifying the above equation, we get,
$a{{m}^{2}}{{b}^{2}}-{{b}^{2}}m\left( am+an \right)+{{a}^{2}}{{\left( m+n \right)}^{2}}c=0$
Or, $a{{m}^{2}}{{b}^{2}}-{{b}^{2}}{{m}^{2}}a-{{b}^{2}}man+{{a}^{2}}{{\left( m+n \right)}^{2}}c=0$
By canceling the like terms from the above equation, we get,
$-{{b}^{2}}mna+{{a}^{2}}{{\left( m+n \right)}^{2}}c=0$
Or, ${{\left( m+n \right)}^{2}}{{a}^{2}}c=amn{{b}^{2}}$
By canceling ‘a’ from both sides, we get
${{\left( m+n \right)}^{2}}ac=mn{{b}^{2}}$
Hence, option (d) is the right answer.

Note: Students must note that whenever we are given any two quantities in ratio, say a:b then we should take the first quantity as ‘ax’ and second quantity as ‘bx’ as the first step to easily solve the problem. Here, x is any variable whose value is to be found. Also, students should remember that in the quadratic equation, the sum of roots $=\dfrac{-b}{a}$ and product of roots $=\dfrac{c}{a}$. Carefully note the values a, b, and c.