Question

If \[\alpha ,\beta \;\] are the roots of \[a{x^2} + bx + c = 0\;\] then the equation whose roots are \[2 + \alpha ,2 + \beta \;\] is:
A. \[a{x^2} + x\left( {4a - b} \right) + 4a - 2b + c = 0\;\]
B. \[a{x^2} + x\left( {4a - b} \right) + 4a + 2b + c = 0\;\]
C. \[a{x^2} + x\left( {b - 4a} \right) - 4a + 2b + c = 0\;\]
D. \[a{x^2} + x\left( {b - 4a} \right) + 4a - 2b + c = 0\;\]

Answer 1

Hint: In this question, we are given that \[\alpha ,\beta \;\] are the roots of \[a{x^2} + bx + c = 0\;\] and we have to find the equation whose root are \[2 + \alpha ,2 + \beta \;\]. Put any one of the roots of the given equation in the equation. Then, equate the root to the unknown equation to $x$ and put the value in the required equation.

Formula Used:
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 \[a{x^2} + bx + c = 0\;\]
It implies that, $x = \alpha ,\beta $ satisfies the above equation.
Substitute, $x = \alpha $ in the given equation \[a{x^2} + bx + c = 0\;\]
We get,
\[a{\alpha ^2} + b\alpha + c = 0\;\] -------(1)
Now, we are given that \[2 + \alpha ,2 + \beta \;\] are the roots of the unknown equation.
We can say, $2 + \alpha = x$
$ \Rightarrow \alpha = x - 2$
Therefore, put $\alpha = x - 2$ in equation (1)
We get,
\[a{\left( {x - 2} \right)^2} + b\left( {x - 2} \right) + c = 0\;\]
Using the algebraic identity i.e., ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
It implies that,
$a\left( {{x^2} + 4 - 4x} \right) + bx - 2b + c = 0$
On simplifying,
$a{x^2} + 4a - 4ax + bx - 2b + c = 0$
Also, written as
\[a{x^2} + x\left( {b - 4a} \right) + 4a - 2b + c = 0\;\]
Thus, the quadratic equation of the roots \[2 + \alpha \;\] and \[2 + \beta \] is \[a{x^2} + x\left( {b - 4a} \right) + 4a - 2b + c = 0\;\].
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. Also, in this question, we must compute the product and sum of the new roots and express them in terms of the coefficients of the provided equation, $a,b$, and $c$. ${x^2} + Sx + P = 0$ is the quadratic equation with sum of roots $S$ and product of roots $P$. We must remember to add an extra negative sign to the negative sign in the sum of the roots when substituting the sum and product. We must ensure that the denominators of the sum and the product of the roots are the same so that we can cancel the denominator by multiplying them with the denominator throughout the equation.