Question

Find the equation whose roots are $\dfrac{{\sqrt a }}{{\sqrt a \pm \sqrt {a - b} }}$.

Answer 1

There are different kinds of equations; they are linear equations, quadratic equations, and polynomial equations. Linear equations will have one root and the quadratic equation will have 2 roots.
The general form of a quadratic equation can be written as ${x^2} - (\alpha + \beta )x + \alpha \beta = 0$, where $\alpha \& \beta$ are the roots of the equation.

Complete step-by-step solution:
It is given that the roots of the equation are $\dfrac{{\sqrt a }}{{\sqrt a + \sqrt {a - b} }}\& \dfrac{{\sqrt a }}{{\sqrt a - \sqrt {a - b} }}$ that is $\dfrac{{\sqrt a }}{{\sqrt a + \sqrt {a - b} }}\& \dfrac{{\sqrt a }}{{\sqrt a - \sqrt {a - b} }}$.
Since there are two roots then the equation is the quadratic equation. Let $\alpha \& \beta$ be the roots of the quadratic equation.
Therefore, $\alpha = \dfrac{{\sqrt a }}{{\sqrt a + \sqrt {a - b} }}$ and $\beta = \dfrac{{\sqrt a }}{{\sqrt a - \sqrt {a - b} }}$.
Let us simplify the first root now,
$\alpha = \dfrac{{\sqrt a }}{{\sqrt a + \sqrt {a - b} }}$
Doing a conjugate multiplication, we get
$\alpha = \dfrac{{\sqrt a }}{{\sqrt a + \sqrt {a - b} }} \times \dfrac{{\sqrt a - \sqrt {a - b} }}{{\sqrt a - \sqrt {a - b} }}$
On simplifying this we get,
$\alpha = \dfrac{{\sqrt a \left( {\sqrt a - \sqrt {a - b} } \right)}}{{\left( {\sqrt a + \sqrt {a - b} } \right)\left( {\sqrt a - \sqrt {a - b} } \right)}}$
Let’s simplify this further.
$\alpha = \dfrac{{\sqrt a \left( {\sqrt a - \sqrt {a - b} } \right)}}{{a - (a - b)}}$
$\Rightarrow \alpha = \dfrac{{\left( {a - \sqrt {a(a - b)} } \right)}}{b}$
Now let’s simplify the second root.
$\beta = \dfrac{{\sqrt a }}{{\sqrt a - \sqrt {a - b} }}$
Doing conjugate multiplication, we get
$\beta = \dfrac{{\sqrt a }}{{\sqrt a - \sqrt {a - b} }} \times \dfrac{{\sqrt a + \sqrt {a - b} }}{{\sqrt a + \sqrt {a - b} }}$
On simplifying this we get,
$\beta = \dfrac{{\sqrt a \left( {\sqrt a + \sqrt {a - b} } \right)}}{{\left( {\sqrt a - \sqrt {a - b} } \right)\left( {\sqrt a + \sqrt {a - b} } \right)}}$
On further simplification, we get
$\beta = \dfrac{{\sqrt a \left( {\sqrt a + \sqrt {a - b} } \right)}}{{a - (a - b)}}$
Simplifying further,
$\beta = \dfrac{{\left( {a + \sqrt {a(a - b)} } \right)}}{b}$
Thus, we have simplified the roots. Let us now substitute them in the general quadratic equation.
We know that the general form of the quadratic equation is ${x^2} - (\alpha + \beta )x + \alpha \beta = 0$, where $\alpha \& \beta$ are the roots of the equation.
Before that, we need to find the value of $\alpha + \beta$ & $\alpha \beta$.
$\alpha + \beta = \dfrac{{a - \sqrt {a(a - b)} + a + \sqrt {a(a - b)} }}{b}$
Simplifying this will give us,
$\Rightarrow \alpha + \beta = \dfrac{{2a}}{b}$
Now let us find the value of $\alpha \beta$.
$\alpha \beta = \dfrac{{\left( {a - \sqrt {a(a - b)} } \right)}}{b} \times \dfrac{{\left( {a + \sqrt {a(a - b)} } \right)}}{b}$
Let’s simplify it further,
$\alpha \beta = \dfrac{{{a^2} - {a^2} + ab}}{{{b^2}}}$
On simplifying it further, we get
$\alpha \beta = \dfrac{a}{b}$
Thus, we found the value of both $\alpha + \beta$ & $\alpha \beta$. Let us now substitute them in the general quadratic equation.
We have${x^2} - (\alpha + \beta )x + \alpha \beta = 0$.
$\Rightarrow {x^2} - \dfrac{{2a}}{b}x + \dfrac{a}{b} = 0$
Simplifying it further we get,
$\Rightarrow b{x^2} - 2ax + a = 0$
Thus, this is the required equation that is, $b{x^2} - 2ax + a = 0$ is the equation whose roots are $\dfrac{{\sqrt a }}{{\sqrt a \pm \sqrt {a - b} }}$.

Note: If we need to find the equation when the roots are given, we first need to find what type of equation is that. Then we need to simplify the given roots then substitute them in the respective general equation. Simplifying that will give us the required equation.