
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
Answer
513.9k+ 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
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

State the laws of reflection of light

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
