Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Radicals. Click on this to access more such competitive quizzes on Number System.

Radicals Quiz 2

mathematics

 Use the property ${\left( {{a^n}} \right)^m} = {a^{nm}}$.

$\sqrt[n]{{{a^n}}} = a$ if $n$ is odd. $ \\ $ Therefore, $\sqrt[3]{{{{\left( { - 3} \right)}^3}}} =  - 3$ 

 $\dfrac{{\sqrt {x + 1}  + \sqrt x }}{{\sqrt {x + 1}  - \sqrt x }}$ in simplified form is equal to

$\dfrac{{\sqrt {x + 1}  + \sqrt x }}{{\sqrt {x + 1}  - \sqrt x }} \times \dfrac{{\sqrt {x + 1}  + \sqrt x }}{{\sqrt {x + 1}  + \sqrt x }} \\   = \dfrac{{{{\left( {\sqrt {x + 1}  + \sqrt x } \right)}^2}}}{{{{\left( {\sqrt {x + 1} } \right)}^2} - {{\left( {\sqrt x } \right)}^2}}} \\   = \dfrac{{x + 1 + x + 2\sqrt {x\left( {x + 1} \right)} }}{{x + 1 - x}} = 2x + 2\sqrt {x\left( {x + 1} \right)}  + 1 $

 $\dfrac{{\sqrt[3]{{8{a^4}{b^2}{c^7}}}}}{{\sqrt {2{a^3}{b^6}{c^2}} }}$ can be simply expressed as

 Use the properties of the exponents to simplify.

$\dfrac{{\sqrt[3]{{8{a^4}{b^2}{c^7}}}}}{{\sqrt {2{a^3}{b^6}{c^2}} }} = \dfrac{{{{\left( 8 \right)}^{\frac{1}{3}}}{{\left( a \right)}^{\frac{4}{3}}}{{\left( b \right)}^{\frac{2}{3}}}{{\left( c \right)}^{\frac{7}{3}}}}}{{{{\left( 2 \right)}^{\frac{1}{2}}}{{\left( a \right)}^{\frac{3}{2}}}{{\left( b \right)}^{\frac{6}{3}}}{{\left( c \right)}^{\frac{2}{2}}}}} \\   = \dfrac{2}{{\sqrt 2 }}{a^{\frac{4}{3} - \frac{3}{2}}}{b^{\frac{2}{3} - 2}}{c^{\frac{7}{3} - 1}} = \sqrt 2 {a^{ - \frac{1}{6}}}{b^{ - \frac{4}{3}}}{c^{\frac{4}{3}}} \\   = \dfrac{{\sqrt 2 .\sqrt[3]{{{c^4}}}}}{{\sqrt[6]{a}.\sqrt[3]{{{b^4}}}}} $

 Simplify $\dfrac{{\sqrt {x + 4} }}{{\sqrt {x - 1}  - 1}} = 3$ 

 Square both sides of the equation after simplifying.

$\dfrac{{\sqrt {x + 4} }}{{\sqrt {x - 1}  - 1}} = 3 \\   \Rightarrow \sqrt {x + 4}  = 3\sqrt {x - 1}  - 3 \\   \Rightarrow 3\sqrt {x - 1}  - \sqrt {x + 4}  = 3 \\  $ Squaring both sides, $\\ \Rightarrow 9\left( {x - 1} \right) + x + 4 - 6\sqrt {\left( {x - 1} \right)\left( {x + 4} \right)}  = 9 \\   \Rightarrow 10x - 14 = 6\sqrt {\left( {x - 1} \right)\left( {x + 4} \right)}  \Rightarrow 5x - 7 = 3\sqrt {\left( {x - 1} \right)\left( {x + 4} \right)}  \\  $ Again squaring both sides, $ \\   \Rightarrow 25{x^2} + 49 - 70x = 9\left( {x - 1} \right)\left( {x + 4} \right) = 9{x^2} - 36 + 27x \\   \Rightarrow 16{x^2} - 97x + 85 = 0 \\   \Rightarrow x = \dfrac{{97 \pm \sqrt {{{97}^2} - 4 \times 85 \times 16} }}{{32}} = \dfrac{{97 \pm \sqrt {9409 - 5440} }}{{32}} = \dfrac{{97 \pm \sqrt {3969} }}{{32}} = \dfrac{{97 \pm 63}}{{32}} \\   \Rightarrow x = \dfrac{{160}}{{32}}\;\;or\;\;\dfrac{{34}}{{32}} = 5\;\;or\;\;\dfrac{{34}}{{32}} \\  $ $x = \dfrac{{34}}{{32}}$ will not satisfy the given equation. $ \\ $ Thus $x = 5$ is the solution.

 If the system of equations $\dfrac{{\sqrt {x - 3}  + \sqrt x }}{{\sqrt {x + 4} }}$ and $\sqrt {3x}  + k$ has a real solution for any real number $k$, then the minimum value of $k$ is

 The minimum value of $k$ will depend on the minimum value of the given function.

$\dfrac{{\sqrt {x - 3}  + \sqrt x }}{{\sqrt {x + 4} }}$ and $\sqrt {3x}  + k$ has real solution. $ \\ $ Therefore, $\dfrac{{\sqrt {x + 4}  + \sqrt x }}{{\sqrt {x + 4}  - \sqrt x }} = x + k$ $ \\ $ Rationalizing the LHS, $ \\  \Rightarrow \dfrac{{\sqrt {x + 4}  + \sqrt x }}{{\sqrt {x + 4}  - \sqrt x }} \times \dfrac{{\sqrt {x + 4}  + \sqrt x }}{{\sqrt {x + 4}  + \sqrt x }} = x + k \\   \Rightarrow \dfrac{{\left( {x + 4} \right) + 2\sqrt {x\left( {x + 4} \right)}  + x}}{4} = x + k \\  $ The domain of $\dfrac{{\sqrt {x + 4}  + \sqrt x }}{{\sqrt {x + 4}  - \sqrt x }}$ is $\left[ {0,\infty } \right)$. $ \\ $ The function $\dfrac{{\sqrt {x + 4}  + \sqrt x }}{{\sqrt {x + 4}  - \sqrt x }}$ is a monotone increasing function. Therefore it attains minimum value at $x = 0$. Thus, $ \\ {\left[ {x + {k_{\min }}} \right]_{x = 0}} = {\left[ {\dfrac{{\sqrt {x + 4}  + \sqrt x }}{{\sqrt {x + 4}  - \sqrt x }}} \right]_{x = 0}} \\   \Rightarrow {k_{\min }} = \dfrac{{\sqrt {0 + 4}  + \sqrt 0 }}{{\sqrt {0 + 4}  - \sqrt 0 }} = 1 $ 