Answer
Verified
478.2k+ 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
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE