Question

If the roots of \[a{x^2} + bx + c = 0\] are in the ratio \[m:n\] , then
A. \[mn{a^2} = (m + n){c^2}\]
B. \[mn{b^2} = (m + n)ac\]
C. \[mn{b^2} = {(m + n)^2}ac\]
D. none of these

Answer 1

Hint:
Since the roots of the given quadratic equation are in the ratio \[m:n\] , we can assume the actual roots to be mx and nx. After this we will find sum of the roots and product of the roots.We would have to square the equation of sum of roots so that the degree of x in sum of roots and product of roots can be equal, because after this we will divide sum of roots equation by product of roots equation in order to get our final answer.

Complete step by step solution:
Let us assume roots of the given quadratic equation \[a{x^2} + bx + c = 0\] be \[\alpha \] and \[\beta \]
According to the question,we get
 \[ \Rightarrow \alpha : \beta = m:n\] … (1)

If we assume the common term in \[\alpha \] and \[\beta \] as \[x\], we can say that the actual values of the roots are
 \[ \Rightarrow \alpha = m \times x\]
 \[ \Rightarrow \beta = n \times x\]

Now,the sum of roots is given by
 \[ \Rightarrow \alpha + \beta = \dfrac{{ - b}}{a}\]

Put \[\alpha = m \times x\] and \[\beta = n \times x\] , we get
 \[ \Rightarrow mx + nx = \dfrac{{ - b}}{a}\]

Taking x common from RHS, we get
 \[ \Rightarrow x \times (m + n) = \dfrac{{ - b}}{a}\]

Squaring the equation, we get
 \[ \Rightarrow {x^2}{(m + n)^2} = \dfrac{{{{( - b)}^2}}}{{{a^2}}}\] … (2)

Now, the product of roots is given by
 \[ \Rightarrow \alpha \times \beta = \dfrac{c}{a}\]

Put \[\alpha = m \times x\] and \[\beta = n \times x\] , we get
 \[ \Rightarrow xm \times xn = \dfrac{c}{a}\]
 \[ \Rightarrow {x^2} \times m \times n = \dfrac{c}{a}\] … (3)

Now, in order to reach the answer we need to form an equation which is free from x
Hence, divide (2) by (3), we get
 \[ \Rightarrow \dfrac{{{x^2}{{(m + n)}^2}}}{{{x^2} \times m \times n}} = \dfrac{{\dfrac{{{{( - b)}^2}}}{{{a^2}}}}}{{\dfrac{c}{a}}}\]

Hence, cancelling \[{x^2}\] and further simplifying the equation, we get
 \[ \Rightarrow \dfrac{{{{(m + n)}^2}}}{{m \times n}} = \dfrac{{{b^2}}}{{ac}}\]

After cross multiplying, we get
 \[ \Rightarrow ac{(m + n)^2} = mn{b^2}\] … (4)

Swapping LHS by RHS in (4), we get
 \[ \Rightarrow mn{b^2} = ac{(m + n)^2}\]

Hence, the final answer is C.

Note:
Do not forget to square the equation of sum of roots, it is the most important part of the question. We square this equation so that when we divide the sum of roots equation by product of roots equation, we can eliminate the common part of the roots to get to the answer. Squaring the sum of roots equation is the MOST important part of this question.