Which of the following is not a quadratic surd?

3^{\frac{1}{2}}

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\sqrt[2]{3}

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\sqrt[3]{2}

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25^{\frac{1}{4}}

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Solution

The correct option is C $\sqrt[3]{2}$
Quadratic surds are the surds whose order is '2'
$3^{\frac{1}{2}}$ and $\sqrt[2]{3}$ has order '2'
$\sqrt[3]{2}$ has order 3
Fourth option $25^{\frac{1}{4}}$ can be written as
$25^{\frac{1}{4}}$ = $(5^{2})^{\frac{1}{4}}$ = $5^{\frac{1}{2}}$ which is again a quadratic surd
Hence only option third $\sqrt[3]{2}$ is not a quadratic surd

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Similar questions

Which of the following are not quadratic polynomials?
(i) $3 + 4 x - 7 x^{2}$

(ii) $8 x^{2} - 15$

(iii) 6x – 15

(iv) $4 x^{3} - 3 x$

(v) $x^{2} - x + 1$

2 (x - 1)^{2} = 4 x^{2} - 2 x + 1

2 x - x^{2} = x^{2} + 5

{(\sqrt{2} x + \sqrt{3})}^{3} = 3 x^{2} - 5 x

(x^{2} + 2 x)^{2} = x^{4} + 3 + 4 x^{2}

Which of the following are not quadratic polynomials?
(i) $3 + 4 x - 7 x^{2}$

(ii) $8 x^{2} - 15$

(iii) 6x – 15

(iv) $4 x^{3} - 3 x$

Which of the following is a quadratic polynomial ?
(a) $x + 4$
(b) $x^{3} + x$
(c) $x^{3} + 2 x + 6$
(d) $x^{2} + 5 x + 4$

(x + 2)^{3} = x^{3} - 4

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