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**Hint:**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.

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