# 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

•

Total views: 305.1k

•

Views today: 3.83k

Answer

Verified

305.1k+ views

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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE