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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) \[\dfrac{{\sqrt a - \sqrt {a - b} }}{{\sqrt b }}\]

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Last updated date: 18th Jun 2024
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Answer
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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.