Answer

Verified

448.2k+ 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

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

How many crores make 10 million class 7 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

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

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths