Question

If one root of the quadratic equation $p{x^2} + qx + r = 0$ ($p \ne 0$) is a surd $\dfrac{{\sqrt a }}{{\sqrt a + \sqrt {a - b} }}$ , where $p$, $q$,$r$,$a$, $b$are all rational then the other root is :
A) $\dfrac{{\sqrt b }}{{\sqrt a + \sqrt {a - b} }}$
B) $a + \dfrac{{\sqrt {a\left( {a - b} \right)} }}{b}$
C) $\dfrac{{a + \sqrt {a\left( {a - b} \right)} }}{b}$
D) $\dfrac{{\sqrt a - \sqrt {a - b} }}{{\sqrt b }}$

Answer 1

Here, we will use the formula for finding the roots of any quadratic equation $a{x^2} + bx + c = 0$ which is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where $a$ is the coefficient of ${x^2}$, $b$ is the coefficient of $x$ and $c$ is the constant term.

Complete step-by-step answer:
Step 1: It is given that one root of the quadratic equation $p{x^2} + qx + r = 0$, $p \ne 0$ is $\dfrac{{\sqrt a }}{{\sqrt a + \sqrt {a - b} }}$ and we have to find the second one.
Let the roots are $\alpha$ and $\beta$ .
$\Rightarrow \alpha = \dfrac{{\sqrt a }}{{\sqrt a + \sqrt {a - b} }}$ ….. (1)
Step 2: In the above expression (1), by doing multiplication and division with $\dfrac{{\sqrt a - \sqrt a - b}}{{\sqrt a - \sqrt {a - b} }}$ , we get:
$\Rightarrow \alpha = \dfrac{{\sqrt a }}{{\sqrt a + \sqrt {a - b} }} \times \dfrac{{\sqrt a - \sqrt a - b}}{{\sqrt a - \sqrt {a - b} }}$
By doing multiplication of the numerators and denominators, we get:
$\Rightarrow \alpha = \dfrac{{{{\left( {\sqrt a } \right)}^2} - \sqrt a \left( {\sqrt a - b} \right)}}{{{{\left( {\sqrt a } \right)}^2} - {{\left( {\sqrt {a - b} } \right)}^2}}}$
Now, as we know that ${\left( {\sqrt a } \right)^2} = a$ , and $\sqrt a \left( {\sqrt a - b} \right) = \sqrt {{a^2} - ab}$ , by substituting these values in the above expression $\alpha = \dfrac{{{{\left( {\sqrt a } \right)}^2} - \sqrt a \left( {\sqrt a - b} \right)}}{{{{\left( {\sqrt a } \right)}^2} - {{\left( {\sqrt {a - b} } \right)}^2}}}$ , we get:
$\Rightarrow \alpha = \dfrac{{a - \sqrt {{a^2} - ab} }}{{a - \left( {a - b} \right)}}$
By simplifying the terms in the numerator and denominator, we get:
$\Rightarrow \alpha = \dfrac{{a - \sqrt {a\left( {a - b} \right)} }}{b}$
Step 3: Now, if one root of the equation is $\alpha = \dfrac{{a - \sqrt {a\left( {a - b} \right)} }}{b}$ , then the second root will be as below:
$\Rightarrow \beta = \dfrac{{a + \sqrt {a\left( {a - b} \right)} }}{b}$ ($\because$ the roots of any quadratic equation are of the form $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and the only term in the square root will have the sign reversed)

Option C is the correct answer.

Note: Students should remember that roots for any quadratic equation $a{x^2} + bx + c = 0$ , will be equal to $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ as proved below:
For the equation $a{x^2} + bx + c = 0$ , by bringing $c$ into the RHS side, we get:
$\Rightarrow a{x^2} + bx = - c$ ………… (1)
By dividing the above equation (1) with $a$, we get:
$\Rightarrow {x^2} + \dfrac{{bx}}{a} = \dfrac{{ - c}}{a}$
Now, by adding ${\left( {\dfrac{b}{{2a}}} \right)^2}$ on both the sides of the equation ${x^2} + \dfrac{{bx}}{a} = \dfrac{{ - c}}{a}$ , we get:
$\Rightarrow {x^2} + \dfrac{{bx}}{a} + {\left( {\dfrac{b}{{2a}}} \right)^2} = \dfrac{{ - c}}{a} + {\left( {\dfrac{b}{{2a}}} \right)^2}$
By using the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ in the above equation, we get:
$\Rightarrow {\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{ - c}}{a} + \dfrac{{{b^2}}}{{4{a^2}}}$
By taking root on both of the sides of the equation ${\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{ - c}}{a} + \dfrac{{{b^2}}}{{4{a^2}}}$ , we get:
$\Rightarrow x + \dfrac{b}{{2a}} = \pm \sqrt {\dfrac{{ - c}}{a} + \dfrac{{{b^2}}}{{4{a^2}}}}$ …………….. (2)
Now by solving the above equation (2) for $x$, we have:
$\Rightarrow x = - \dfrac{b}{{2a}} \pm \sqrt {\dfrac{{ - c}}{a} + \dfrac{{{b^2}}}{{4{a^2}}}}$
By multiplying $4a$ with the term $\dfrac{{ - c}}{a}$ , for making the denominator the same inside the square root, we get:
$\Rightarrow x = - \dfrac{b}{{2a}} \pm \sqrt {\dfrac{{ - 4ac}}{{4{a^2}}} + \dfrac{{{b^2}}}{{4{a^2}}}}$
By taking LCM inside the square root, we get:
$\Rightarrow x = - \dfrac{b}{{2a}} \pm \sqrt {\dfrac{{{b^2} - 4ac}}{{4{a^2}}}}$
We can write the above equation as below:
$\Rightarrow x = - \dfrac{b}{{2a}} \pm \dfrac{{\sqrt {{b^2} - 4ac} }}{{\sqrt {4{a^2}} }}$
By substituting the values of $\sqrt {4{a^2}} = 2a$ in the above equation $x = - \dfrac{b}{{2a}} \pm \dfrac{{\sqrt {{b^2} - 4ac} }}{{\sqrt {4{a^2}} }}$ , we get:
$\Rightarrow x = - \dfrac{b}{{2a}} \pm \dfrac{{\sqrt {{b^2} - 4ac} }}{{2a}}$
By taking LCM in the RHS side in the above equation $x = - \dfrac{b}{{2a}} \pm \dfrac{{\sqrt {{b^2} - 4ac} }}{{2a}}$, we get:
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Hence, the roots of a quadratic equation are given by, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Also, you should remember that if,
${b^2} - 4ac < 0$, There are no real roots.
${b^2} - 4ac = 0$, There is one real root.
${b^2} - 4ac > 0$, There are two real roots.