Question

If equation $a{{x}^{2}}+bx+c\ =\ 0$, $\left( a,b,c\in \text{R,}\ \text{a}\ne \text{0} \right)$ and $3{{x}^{2}}+5x+4\ =\ 0$ have common roots, then $a:b:c$ equals to:
A). $1:2:3$
B). $2:3:4$
C). $4:3:2$
D). $3:5:4$

Answer 1

First find discrimination then say that if one root is same then both the roots have to be same as the roots are complex and conjugate. Then compare the unknown and known equations sum and product of roots.

Complete step-by-step solution:
In the question common equation are given $a{{x}^{2}}+bx+c\ =\ 0$, $\left( a,b,c\in \text{R,}\ \text{a}\ne \text{0} \right)$ and $3{{x}^{2}}+5x+4\ =\ 0$ have common root so we have to find ratio of a, b, c.
Now let’s first analyze the equation $3{{x}^{2}}+5x+4\ =\ 0$
We will find out the discriminant of equation using formula $\text{D}\ \text{=}\ {{\text{b}}^{\text{2}}}-4ac$, if the equation is $a{{x}^{2}}+bx+c\ =\ 0$.
So, the discriminant is
$\text{D}\ \text{=}\ {{5}^{\text{2}}}-4\centerdot 3\centerdot 4$
$=\ 25-48\ =\ -23$
Here, $\text{D} < 0$ , so we can tell that the roots of the equation are imaginary.
Now, if one of the roots is imaginary then we can say that both are imaginary and exist in conjugate pairs.
Now, let’s suppose that one of the roots is $x+iy$ then another root should be $x-iy$.
If $\left( x+iy \right)$ is the common root then $\left( x-iy \right)$should also be the common root between them.
Now, let’s compare the values of sum of the roots and products of roots.
We apply the formula that if the given equation is $a{{x}^{2}}+bx+c\ =\ 0$ and roots are $\alpha ,\beta$ then $\alpha +\beta \ =\ \dfrac{-b}{a}$ and $\alpha \times \beta \ =\ \dfrac{c}{a}$.
In the first equation $3{{x}^{2}}+5x+4\ =\ 0$ we get sum of roots as,
$\alpha +\beta \ =\ \dfrac{-5}{3}$
In the second equation $a{{x}^{2}}+bx+c\ =\ 0$ we get sum of roots as,
$\alpha +\beta \ =\ \dfrac{-b}{a}$
In the first equation sum of the roots were $\dfrac{-5}{3}$ and in the second equation sum of the roots are $\dfrac{-b}{a}$ so, we can equate them as we know that their both the roots are equal.
So, $\dfrac{-b}{a}\ =\ \dfrac{-5}{3}$
Hence $\dfrac{a}{b}\ =\ \dfrac{3}{5}$.
Or $a:b\ =\ 3:5$
Now, we will compare the products of roots.
In the first equation $3{{x}^{2}}+5x+4\ =\ 0$ we get product of roots as,
$\alpha \beta \ =\ \dfrac{4}{3}$
In the second equation $a{{x}^{2}}+bx+c\ =\ 0$ we get product of roots as,
$\alpha \beta \ =\ \dfrac{c}{a}$
As we know that roots are equal so their products will also be equal.
So, $\dfrac{c}{a}\ =\ \dfrac{4}{3}$
Or $c:a\ =\ 4:3$
We know that $\dfrac{a}{b}\ =\ \dfrac{3}{5}$ and $\dfrac{c}{a}\ =\ \dfrac{4}{3}$. So, $\dfrac{c}{b}\ =\ \dfrac{c}{a}\times \dfrac{a}{b}$ which can be written as $\dfrac{c}{b}\ =\ \dfrac{4}{3} \times \dfrac{3}{5}\ =\ \dfrac{4}{5}$
Hence, $a:b\ =\ 3:5$ and $b:c\ =\ 5:4$
So, $a:b:c\ =\ 3:5:4$
Hence, the correct option is (D).

Note: One can also find the common root first and then compare by adding the coefficient of ${{x}^{2}}$, the equations same and then subtracting each other or using formula if equations are ${{a}_{1}}{{x}^{2}}+{{b}_{1}}x+{{c}_{1}}\ =\ 0$ and ${{a}_{2}}{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}\ =\ 0$, common roots be d so,
$\text{d}\ =\ \dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}\ =\ \dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\$