Question

If the roots of $a{x^2} + bx + c = 0$ are \[\alpha ,\beta \] and the roots of \[A{x^2} + Bx + C = 0\] are \[\left( {\alpha - k} \right),\left( {\beta - k} \right)\]. Then $\dfrac{{{B^2} - 4AC}}{{{b^2} - 4ac}}$ is equal to
A. $0$
B. $1$
C. ${\left( {\dfrac{A}{a}} \right)^2}$
D. ${\left( {\dfrac{a}{A}} \right)^2}$

Answer 1

Hint: In this question, we are given two quadratic equations and their roots. Also, we have to find the value of $\dfrac{{{B^2} - 4AC}}{{{b^2} - 4ac}}$. Here, the value we have to calculate is in the form of discriminant $D = {b^2} - 4ac$. We’ll apply the formula of difference of the roots i.e., $\alpha - \beta = \dfrac{{\sqrt {{b^2} - 4ac} }}{a}$ .

Formula Used:
General quadratic equation: – $a{x^2} + bx + c = 0$
Let, the roots of the above quadratic equation be $\alpha $ and $\beta $
Therefore,
Difference of the roots will be $\alpha - \beta = \dfrac{{\sqrt {{b^2} - 4ac} }}{a}$

Complete step by step Solution:
Given that,
\[\alpha ,\beta \] are the roots of the quadratic equation $a{x^2} + bx + c = 0$
Using the formula of the difference of roots,
We get,
$\alpha - \beta = \dfrac{{\sqrt {{b^2} - 4ac} }}{a}$ ---------(1)
Also, \[\left( {\alpha - k} \right),\left( {\beta - k} \right)\] are the roots of the quadratic equation \[A{x^2} + Bx + C = 0\]
Again, using the difference formula of roots.
We get,
$\left( {\alpha - k} \right) - \left( {\beta - k} \right) = \dfrac{{\sqrt {{B^2} - 4AC} }}{A}$
On solving the above equation,
We get $\alpha - \beta = \dfrac{{\sqrt {{B^2} - 4AC} }}{A}$
From equation (1), substitute the value of $\alpha - \beta = \dfrac{{\sqrt {{b^2} - 4ac} }}{a}$ in the above equation.
It implies that,
$\dfrac{{\sqrt {{b^2} - 4ac} }}{a} = \dfrac{{\sqrt {{B^2} - 4AC} }}{A}$
Also written as,
$\dfrac{{\sqrt {{B^2} - 4AC} }}{{{b^2} - 4ac}} = \dfrac{A}{a}$
To remove the square root from the numerator and denominator. Squaring both sides of the above equation.
${\left( {\dfrac{{\sqrt {{B^2} - 4AC} }}{{{b^2} - 4ac}}} \right)^2} = {\left( {\dfrac{A}{a}} \right)^2}$
Thus, the value of $\dfrac{{{B^2} - 4AC}}{{{b^2} - 4ac}}$ is equal to ${\left( {\dfrac{A}{a}} \right)^2}$.

Hence, the correct option is (C).

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.