Question

If the ratio of the roots of $a{x^2} + bx + c = 0$ is same as the roots of $p{x^2} + qx + r = 0$ then:
A. $\dfrac{b}{{ac}} = \dfrac{q}{{pr}}$
B. $\dfrac{{{b^2}}}{{ac}} = \dfrac{{{q^2}}}{{pr}}$
C. $\dfrac{{2b}}{{ac}} = \dfrac{{{q^2}}}{{pr}}$
D. None of these

Answer 1

Hint: In this question, we are given two quadratic equation i.e., $a{x^2} + bx + c = 0$ and $p{x^2} + qx + r = 0$. We have to find the ratio of their roots where the ratio of the roots of the equations is equal. Firstly, calculate the sum and the product of roots. Then, write the ratio of roots in two ways and add them together.

Formula Used:
General quadratic equation: – \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Sum of roots, $\alpha + \beta = \dfrac{{ - B}}{A}$
Product of roots, $\alpha \beta = \dfrac{C}{A}$

Complete step by step Solution:
Given that,
There are two quadratic equations $a{x^2} + bx + c = 0$ and $p{x^2} + qx + r = 0$.
Let, \[\alpha ,\beta \] are the roots of $a{x^2} + bx + c = 0$
And \[\gamma ,\delta \] are the roots of $p{x^2} + qx + r = 0$
Now, using the formula of sum and product of the roots in the equations
We get,
For $a{x^2} + bx + c = 0$,
Sum, $\alpha + \beta = \dfrac{{ - b}}{a}$ and $\alpha \beta = \dfrac{c}{a}$ -------(1)
For $p{x^2} + qx + r = 0$.
Sum, $\gamma + \delta = \dfrac{{ - q}}{p}$ and $\gamma \delta = \dfrac{r}{p}$ -------(2)
According to the question,
The ratio of the root is equal. It implies that,
\[\dfrac{\alpha }{\beta } = \dfrac{\gamma }{\delta }\] ---------(3)
And \[\dfrac{\beta }{\alpha } = \dfrac{\delta }{\gamma }\] ---------(4)
Adding equations (3) and (4),
\[\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha } = \dfrac{\gamma }{\delta } + \dfrac{\delta }{\gamma }\]
On simplifying we get,
\[\dfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} = \dfrac{{{\gamma ^2} + {\delta ^2}}}{{\gamma \delta }}\]
Adding $2$ on both sides,
\[\dfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} + 2 = \dfrac{{{\gamma ^2} + {\delta ^2}}}{{\gamma \delta }} + 2\]
\[\dfrac{{{\alpha ^2} + {\beta ^2} + 2\alpha \beta }}{{\alpha \beta }} = \dfrac{{{\gamma ^2} + {\delta ^2} + 2\gamma \delta }}{{\gamma \delta }}\]
(Converting the numerator using the algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$)
\[\dfrac{{{{\left( {\alpha + \beta } \right)}^2}}}{{\alpha \beta }} = \dfrac{{{{\left( {\gamma + \delta } \right)}^2}}}{{\gamma \delta }}\]
Substituting the values from equations (1) and (2),
\[\dfrac{{{{\left( {\dfrac{{ - b}}{a}} \right)}^2}}}{{\left( {\dfrac{c}{a}} \right)}} = \dfrac{{{{\left( {\dfrac{{ - q}}{p}} \right)}^2}}}{{\left( {\dfrac{r}{p}} \right)}}\]
On solving it further,
\[\dfrac{{{b^2}}}{{ac}} = \dfrac{{{q^2}}}{{pr}}\]

Hence, the correct option is (B).

Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variable $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x - $ coordinates of the function's $x - $ intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.