Question

Let $\alpha $ and $\beta $ be the roots of the equation ${x^2} + 2px - 2{q^2}$ where $p$ and $q$ are rational and ${p^2} + {q^2}$ is not a perfect square, form the quadratic equation whose one root is $\alpha + \beta + \sqrt {{\alpha ^2} + {\beta ^2}} $?

Answer 1

Hint: Here in this question, we have to find the quadratic equation by using a given condition. For this, we need to find the sum and product of the given roots of the equation ${x^2} + 2px - 2{q^2}$ and on further substitute simplify in the root $\alpha + \beta + \sqrt {{\alpha ^2} + {\beta ^2}} $ to get the required solution.

Complete step by step answer:
A quadratic equation in x is an equation that can be written in the standard form $a{x^2} + bx + c = 0$, where a, b and c are real numbers and $a \ne 0$. ‘a’ represents the numerical coefficient of ${x^2}$, ‘b’ represents the numerical coefficient of $x$, and ‘c’ represents the constant numerical term. For every quadratic equation, there can be one or more than one roots or solution. These are called the roots of the quadratic equation.For a quadratic equation,
$a{x^2} + bx + c = 0$
The sum of its roots $ = - \dfrac{b}{a}$ and the product of its roots $ = \dfrac{c}{a}$.

Consider the given question, given the $\alpha $ and $\beta $ be the roots of the equation ${x^2} + 2px - 2{q^2}$.
Here $a = 1$, $b = 2p$ and $c = - 2{q^2}$.
Sum of their roots $\alpha + \beta = - \dfrac{{2p}}{1} = - 2p$ ----- (1)
And their product $\alpha \beta = \dfrac{{ - 2{q^2}}}{1} = - 2{q^2}$ ----- (2)
Let us consider the root $\alpha + \beta + \sqrt {{\alpha ^2} + {\beta ^2}} $
Add and subtract $2\alpha \beta $ inside the square root.
$ \Rightarrow \,\,\,\alpha + \beta + \sqrt {{\alpha ^2} + {\beta ^2} + 2\alpha \beta - 2\alpha \beta } $

By the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, then
$ \Rightarrow \,\,\,\alpha + \beta + \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } $
By (1) and (2)
$ \Rightarrow \,\,\left( { - 2p} \right) + \sqrt {{{\left( { - 2p} \right)}^2} - \left( { - 2{q^2}} \right)} $
$ \Rightarrow \,\, - 2p + \sqrt {4{p^2} + 4{q^2}} $
$ \Rightarrow \,\, - 2p + \sqrt {4\left( {{p^2} + {q^2}} \right)} $
$ \Rightarrow \,\, - 2p + 2\sqrt {\left( {{p^2} + {q^2}} \right)} $

Given, as $p$ and $q$ are rational and $\sqrt {{p^2} + {q^2}} $ is not rational, we must have the other root of the desired equation as its conjugate i.e., $ - 2p - 2\sqrt {\left( {{p^2} + {q^2}} \right)} $.
The two roots $ - 2p + 2\sqrt {\left( {{p^2} + {q^2}} \right)} $ and $ - 2p - 2\sqrt {\left( {{p^2} + {q^2}} \right)} $, then
Sum of roots is:
$ \Rightarrow \,\, - 2p + 2\sqrt {\left( {{p^2} + {q^2}} \right)} + \left( { - 2p - 2\sqrt {\left( {{p^2} + {q^2}} \right)} } \right)$
$ \Rightarrow \,\, - 2p + 2\sqrt {\left( {{p^2} + {q^2}} \right)} - 2p - 2\sqrt {\left( {{p^2} + {q^2}} \right)} $
On simplification, we get
$ \Rightarrow \,\, - 4p$

Therefore, the sum of roots is $ - 4p$.And product of root is:
$ \Rightarrow \,\, - 2p + 2\sqrt {\left( {{p^2} + {q^2}} \right)} \,\, \cdot \left( { - 2p - 2\sqrt {\left( {{p^2} + {q^2}} \right)} } \right)$
$ \Rightarrow \,\, - 2p \cdot \left( { - 2p - 2\sqrt {\left( {{p^2} + {q^2}} \right)} } \right) + 2\sqrt {\left( {{p^2} + {q^2}} \right)} \cdot \left( { - 2p - 2\sqrt {\left( {{p^2} + {q^2}} \right)} } \right)$
$ \Rightarrow \,\,4{p^2} + 4p\sqrt {\left( {{p^2} + {q^2}} \right)} - 4p\sqrt {\left( {{p^2} + {q^2}} \right)} - 4{\left( {\sqrt {\left( {{p^2} + {q^2}} \right)} } \right)^2}$
$ \Rightarrow \,\,4{p^2} - 4\left( {{p^2} + {q^2}} \right)$
$ \Rightarrow \,\,4{p^2} - 4{p^2} - 4{q^2}$
On simplification, we get
$ \therefore \,\, - 4{q^2}$
Therefore, the product of roots is $ - 4{q^2}$.

Hence, the required quadratic equation is: ${x^2} + 4px - 4{q^2} = 0$.

Note: The general form of quadratic equation is represented as $a{x^2} + bx + c = 0$ it represents the polynomial equation whose variable term has degree 2. To simplify we use the simple arithmetic operations like addition, subtraction, multiplication and division and we should know about the tables of multiplication.