Question

The roots of the quadratic equation \[\left( {a + b - 2c} \right){x^2} - \left( {2a - b - c} \right)x + \left( {a - 2b + c} \right) = 0\;\] are -
A. $\left( {a + b + c} \right)$ and $\left( {a - b - c} \right)$
B. $\dfrac{1}{2}$ and $a - 2b + c$
C. $a - 2b + c$ and $\dfrac{1}{{a + b - 2c}}$
D. None of the above

Answer 1

Hint: In this question, we are given a quadratic equation \[\left( {a + b - 2c} \right){x^2} - \left( {2a - b - c} \right)x + \left( {a - 2b + c} \right) = 0\;\] and we have to calculate the roots of the equation. Firstly, put $x = 1$. You’ll get the one root and then let the other root be $\alpha $. In last, apply the formula of the product of roots $\alpha \beta = \dfrac{C}{A}$ where the equation is \[A{x^2} + Bx + C = 0\].

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,
\[\left( {a + b - 2c} \right){x^2} - \left( {2a - b - c} \right)x + \left( {a - 2b + c} \right) = 0\;\], a quadratic equation
At $x = 1$,
Taking the right-hand side of the equation,
\[ = \left( {a + b - 2c} \right){\left( 1 \right)^2} - \left( {2a - b - c} \right)\left( 1 \right) + \left( {a - 2b + c} \right)\;\]
$ = a + b - 2c - 2a + b + c + a - 2b + c$
Cancel the like terms,
We get, $ = 0$ (Left hand side)
It implies that, $x = 1$ satisfies the given equation.
Thus, $1$ is the root of the equation \[\left( {a + b - 2c} \right){x^2} - \left( {2a - b - c} \right)x + \left( {a - 2b + c} \right) = 0\;\]
Now, let the other root of the equation be $\alpha $
Compare the given equation \[\left( {a + b - 2c} \right){x^2} - \left( {2a - b - c} \right)x + \left( {a - 2b + c} \right) = 0\;\] with the general quadratic equation \[A{x^2} + Bx + C = 0\],
We get, $A = \left( {a + b - 2c} \right)$, $B = - \left( {2a - b - c} \right)$, and $C = \left( {a - 2b + c} \right)$
Applying the formula of the product of the roots,
It will be, $1 \times \alpha = \dfrac{{a - 2b + c}}{{a + b - 2c}}$
$\alpha = \dfrac{{a - 2b + c}}{{a + b - 2c}}$
Therefore, the roots of the given equation are $1$ and $\dfrac{{a - 2b + c}}{{a + b - 2c}}$.

Hence, the correct option is (D).

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.