Question

If $\alpha$, $\beta$ are roots of $A{x^2} + Bx + C = 0$ and ${\alpha ^2}$, ${\beta ^2}$ are roots of ${x^2} + px + q = 0$, the $p$ is equal to
A. $\dfrac{{{B^2} - 2AC}}{{{A^2}}}$
B. $\dfrac{{2AC - {B^2}}}{{{A^2}}}$
C. $\dfrac{{{B^2} - 4AC}}{{{A^2}}}$
D. $\dfrac{{4AC - {B^2}}}{{{A^2}}}$

Answer 1

Hint: In this question, we are given two quadratic equations and their roots. We have to calculate the value of $p$. Firstly, apply the formula of sum and product of the quadratic equation i.e., $\alpha + \beta = \dfrac{{ - b}}{a}$ and $\alpha \beta = \dfrac{c}{a}$ where the equation is \[a{x^2} + bx + c = 0\]. Then, apply the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ to solve it further.

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, $\alpha \beta = \dfrac{c}{a}$
Algebraic identity –
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

Complete step by step solution:
Given that,
$\alpha $, $\beta $ are the roots of $A{x^2} + Bx + C = 0$
And ${\alpha ^2}$, ${\beta ^2}$ are the roots of ${x^2} + px + q = 0$
As we know that, if $\alpha $, $\beta $ are the roots of the equation $a{x^2} + bx + c = 0$ then the sum of the roots and the product of the roots are $\dfrac{{ - b}}{a}$, $\dfrac{c}{a}$ respectively
It implies that,
For $A{x^2} + Bx + C = 0$,
$\alpha + \beta = \dfrac{{ - B}}{A}$ and $\alpha \beta = \dfrac{C}{A}$ ----------- (1)
And for ${x^2} + px + q = 0$,
${\alpha ^2} + {\beta ^2} = - p$ and ${\alpha ^2}{\beta ^2} = q$ ----------- (2)
As we know that, $\alpha + \beta = \dfrac{{ - B}}{A}$
Squaring both the sides of the above equation
We get, ${\left( {\alpha + \beta } \right)^2} = {\left( {\dfrac{{ - B}}{A}} \right)^2}$
Using the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ in equation (1),
It implies that, ${\alpha ^2} + {\beta ^2} + 2\alpha \beta = \dfrac{{{B^2}}}{{{A^2}}}$
Using equation (1), it will be ${\alpha ^2} + {\beta ^2} + 2\left( {\dfrac{C}{A}} \right) = \dfrac{{{B^2}}}{{{A^2}}}$
${\alpha ^2} + {\beta ^2} = \dfrac{{{B^2}}}{{{A^2}}} - \dfrac{{2C}}{A}$
Here, the L.C.M. of ${A^2}$ and $A$ is ${A^2}$.
On solving, we get ${\alpha ^2} + {\beta ^2} = \dfrac{{{B^2} - 2AC}}{{{A^2}}}$
Using equation (2) i.e., ${\alpha ^2} + {\beta ^2} = - p$
We get, $ - p = \dfrac{{{B^2} - 2AC}}{{{A^2}}}$
Also written as $p = \dfrac{{2AC - {B^2}}}{{{A^2}}}$

Hence, option (B) is the correct answer i.e., $\dfrac{{2AC - {B^2}}}{{{A^2}}}$.

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.