Question

If the roots of the equation, $a{x^2} + bx + c = 0$, are of the form \[\dfrac{\alpha }{{\left( {\alpha - 1} \right)}}\;\] and \[\dfrac{{\left( {\alpha + 1} \right)}}{\alpha }\], then the value of \[{\left( {a + b + c} \right)^2}\] is
A. $2{b^2} - ac$
B. ${b^2} - 2ac$
C. ${b^2} - 4ac$
D. $4{b^2} - 2ac$

Answer 1

Hint:In this question, we are given a quadratic equation $a{x^2} + bx + c = 0$ whose roots are \[\dfrac{\alpha }{{\left( {\alpha - 1} \right)}}\;\] and \[\dfrac{{\left( {\alpha + 1} \right)}}{\alpha }\]. We have to find the value of \[{\left( {a + b + c} \right)^2}\]. Firstly, apply the formula of sum and the product of roots. Using the product formula calculates the value of $\alpha $ and put the required value in any of the roots. In the last put the value of the root in the given equation and 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, $\alpha \beta = \dfrac{C}{A}$

Complete step by step Solution:
Given that,
$a{x^2} + bx + c = 0$, a quadratic equation
The roots of the above equation are \[\dfrac{\alpha }{{\left( {\alpha - 1} \right)}}\;\] and \[\dfrac{{\left( {\alpha + 1} \right)}}{\alpha }\]
Now, using the formula of sum and product
We get,
Sum, $\dfrac{\alpha }{{\left( {\alpha - 1} \right)}}\; + \dfrac{{\left( {\alpha + 1} \right)}}{\alpha } = \dfrac{{ - b}}{a}$
And product will be $\dfrac{\alpha }{{\left( {\alpha - 1} \right)}}\; \times \dfrac{{\left( {\alpha + 1} \right)}}{\alpha } = \dfrac{c}{a}$
Cancel the like terms of numerator and denominator,
$\dfrac{{\left( {\alpha + 1} \right)}}{{\left( {\alpha - 1} \right)}} = \dfrac{c}{a}$
Cross – multiplying both the side,
$a\alpha + a = c\alpha - c$
$c\alpha - a\alpha = a + c$
$\alpha = \dfrac{{a + c}}{{c - a}} - - - - - - - \left( 1 \right)$
As we know,
In a quadratic equation, roots are the values of the variable.
It implies that, the values of $x$ are \[\dfrac{\alpha }{{\left( {\alpha - 1} \right)}}\;\] and \[\dfrac{{\left( {\alpha + 1} \right)}}{\alpha }\] . Also, they will satisfy the quadratic equation.
Let’s calculate the value of \[\dfrac{\alpha }{{\left( {\alpha - 1} \right)}}\;\]
\[\dfrac{\alpha }{{\left( {\alpha - 1} \right)}}\; = \dfrac{{\left( {\dfrac{{a + c}}{{c - a}}} \right)}}{{\left( {\dfrac{{a + c}}{{c - a}} - 1} \right)}}\;\]
On simplifying, we get \[\dfrac{\alpha }{{\left( {\alpha - 1} \right)}}\; = \dfrac{{a + c}}{{2a}}\;\]
Put \[x = \dfrac{{a + c}}{{2a}}\;\] in the given equation $a{x^2} + bx + c = 0$
It implies that,
$a{\left( {\dfrac{{a + c}}{{2a}}} \right)^2} + b\left( {\dfrac{{a + c}}{{2a}}} \right) + c = 0$
Using the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
$a\left( {\dfrac{{{a^2} + {c^2} + 2ac}}{{4{a^2}}}} \right) + b\left( {\dfrac{{a + c}}{{2a}}} \right) + c = 0$
$\left( {\dfrac{{{a^2} + {c^2} + 2ac}}{{4a}}} \right) + \left( {\dfrac{{ab + bc}}{{2a}}} \right) + c = 0$
Taking the denominator as $4a$,
$\left( {\dfrac{{{a^2} + {c^2} + 2ac}}{{4a}}} \right) + \left( {\dfrac{{ab + bc}}{{2a}}} \right) + c = 0$
$ \Rightarrow {a^2} + {c^2} + 2ac + 2ab + 2bc + 4ac = 0$
Add ${b^2}$ to both the sides of the above equation,
$ \Rightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac = {b^2} - 4ac$
Right side of the equation is in the algebraic identity form i.e., ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac$
Thus, ${\left( {a + b + c} \right)^2} = {b^2} - 4ac$

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.