Answer

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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.

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.

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