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# If the roots of $a{x^2} + bx + c = 0$ are in the ratio $m:n$ thenA. $mn{a^2} = \left( {m + n} \right){c^2}$ B. $mn{b^2} = \left( {m + n} \right)ac$ C. $mn{b^2} = {\left( {m + n} \right)^2}ac$ D. None of these

Last updated date: 16th Jun 2024
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Hint: First find the sum of roots and product of the roots of the given equation. Assume the roots to be my and ny. Then put the value of y from the sum equation to the product of the roots and adjust the equation.

Given the roots of $a{x^2} + bx + c = 0$ are in the ratio $m:n$ then
Let the roots be ‘my’ and ‘ny’.
Now we know that sum of roots=$- \dfrac{{{\text{coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$
And product of roots=$\dfrac{{{\text{constant}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$
On putting the values in the formula we get,
$\Rightarrow my + ny = \dfrac{{ - b}}{a}$
$\Rightarrow y\left( {m + n} \right) = \dfrac{{ - b}}{a}$ --- (i)
And $my \times ny = \dfrac{c}{a}$
$\Rightarrow$ $mn{y^2} = \dfrac{c}{a}$ --- (ii)
From eq. (i) we get,
$\Rightarrow y = \dfrac{{ - b}}{{a\left( {m + n} \right)}}$
On putting the value of y from eq. (i) to (ii) we get,
$\Rightarrow mn{\left( {\dfrac{{ - b}}{{a\left( {m + n} \right)}}} \right)^2} = \dfrac{c}{a}$
On simplifying we get,
$\Rightarrow mn\dfrac{{{{\left( { - b} \right)}^2}}}{{{a^2}{{\left( {m + n} \right)}^2}}} = \dfrac{c}{a}$
On transferring ${a^2}{\left( {m + n} \right)^2}$ on the right side we get,
$\Rightarrow mn \times {b^2} = \dfrac{{c{a^2}{{\left( {m + n} \right)}^2}}}{a}$
On cancelling a from denominator and numerator we get,
$\Rightarrow mn{b^2} = ac{\left( {m + n} \right)^2}$

Hence the correct answer is ‘C’.

Note:: This method can also be solved using the following method-
Assume the roots of the given equation to be $\alpha$ and $\beta$ then according to question,
$\Rightarrow \dfrac{\alpha }{\beta } = \dfrac{m}{n}$ -- (I)
Using Componendo, we get
$\Rightarrow \dfrac{{\alpha + \beta }}{\beta } = \dfrac{{m + n}}{n}$ $\Rightarrow \dfrac{{\alpha + \beta }}{{m + n}} = \dfrac{\beta }{n}$ -- (II)
We can also write eq.(I) as $\dfrac{\alpha }{m} = \dfrac{\beta }{n}$
On multiplying$\dfrac{\beta }{n}$ both sides we get $\dfrac{{\alpha \beta }}{{mn}} = {\left( {\dfrac{\beta }{n}} \right)^2}$
$\Rightarrow \sqrt {\dfrac{{\alpha \beta }}{{mn}}} = \dfrac{\beta }{n}$ --- (III)
On equating eq. (II) and (III) we get,
$\Rightarrow \dfrac{{\alpha + \beta }}{{m + n}} = \sqrt {\dfrac{{\alpha \beta }}{{mn}}}$
On solving we get,
$\Rightarrow \dfrac{{{{\left( {\alpha + \beta } \right)}^2}}}{{{{\left( {m + n} \right)}^2}}} = \dfrac{{\alpha \beta }}{{mn}}$
On cross multiplication we get,
$\Rightarrow {\left( {m + n} \right)^2}\alpha \beta = {\left( {\alpha + \beta } \right)^2}mn$
Now we already know the value of the sum of roots and the product of roots, so putting the values in the equation we get,
$\Rightarrow {\left( {m + n} \right)^2}\dfrac{c}{a} = {\left( {\dfrac{{ - b}}{a}} \right)^2}mn$
On simplifying we get,
$\Rightarrow mn{b^2} = ac{\left( {m + n} \right)^2}$

Hence, the correct answer is obtained.