Question

If \[a{(p + q)^2} + 2bpq + c = 0\] and \[a{(p + r)^2} + 2bpr + c = 0\], then $qr = $
A) ${p^2} + \dfrac{c}{a}$
B) ${p^2} + \dfrac{a}{c}$
C) ${p^2} + \dfrac{a}{b}$
D) ${p^2} + \dfrac{b}{a}$

Answer 1

Hint: In this question, we are given two quadratic equations i.e., \[a{(p + q)^2} + 2bpq + c = 0\] and \[a{(p + r)^2} + 2bpr + c = 0\]. We have to find the value of the product of $q$ and $r$. Let, the equation be as $x$ variable where the roots of $x$ will be $q$ and $r$. Then apply the formula of the product of the roots.

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,
\[a{(p + q)^2} + 2bpq + c = 0\] and \[a{(p + r)^2} + 2bpr + c = 0\]
Now, we can see both equations are in the same format with different variables.
We can write it as \[a{(p + x)^2} + 2bpx + c = 0\]
Where $q$ and $r$ are the roots of the above equation.
Now, we’ll solve the required equation by applying the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
So, \[a{(p + x)^2} + 2bpx + c = 0\]
\[ \Rightarrow a\left( {{p^2} + {x^2} + 2px} \right) + 2bpx + c = 0\]
$ \Rightarrow a{p^2} + a{x^2} + 2apx + 2bpx + c = 0$
On solving, we get
$ \Rightarrow a{x^2} + \left( {2ap + 2bp} \right)x + \left( {c + a{p^2}} \right) = 0$
Now, compare the required equation $a{x^2} + \left( {2ap + 2bp} \right)x + \left( {c + a{p^2}} \right) = 0$ with the general quadratic equation i.e., \[A{x^2} + Bx + C = 0\]
It implies that,
$A = a$, $B = 2ap + 2bp$, and $C = c + a{p^2}$
As we know that, $q$ and $r$ are the roots of the equation $a{x^2} + \left( {2ap + 2bp} \right)x + \left( {c + a{p^2}} \right) = 0$
So, using the formula of the product of root $\alpha \beta = \dfrac{C}{A}$
It will be, $qr = \dfrac{{c + a{p^2}}}{a}$
Taking $a$ common from the numerator,
$qr = \dfrac{{a\left( {\dfrac{c}{a} + {p^2}} \right)}}{a}$
Canceling the like terms,
$qr = \dfrac{c}{a} + {p^2}$
Thus, the value of $qr$ is equals to ${p^2} + \dfrac{c}{a}$.

Hence, the correct option is (A).

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.