CAT 2025 Lesson : Surds & Indices - Root of Surds

(a + b)^2 = a^2 + b^2 + 2ab

To find the square root of a quadratic surd, we express the surd in the form of

(a^2 + b^2 + 2ab)

a^2 + b^2

forms the rational part and

2ab

forms the irrational part.

Find the square root of

9 + 4 \sqrt{5}

9 + 4 \sqrt{5} = 9 + 2 \sqrt{20}

= 4 + 5 + 2 \sqrt{{4} \times {5}}

= (2)^2 + (\sqrt{5})^2 + 2 \times 2 \times \sqrt{5}

= (2 + \sqrt{5})^2

\therefore \sqrt{{9 + 4 \sqrt{5}}} = \sqrt{{(2 + \sqrt {5})^2}} = \pm(2 + \sqrt{5})

Answer:

\pm(2 + \sqrt{5})

Find the square root of

15 - 4 \sqrt{14}

(1)

\sqrt{7} - 2\sqrt{2}

(2)

\sqrt{7} - \sqrt{6}

(3)

\sqrt{8} - \sqrt{7}

(4) More than one of the above

Let

x = 15 - 4 \sqrt{14} = 15 - 2 \sqrt{56}

= 7 + 8 - 2 \sqrt{{7} \times {8}}

= (\sqrt{7} - \sqrt{8})^2

\sqrt{x} = \sqrt{(\sqrt{7} - {8})^2} = \pm(\sqrt{7} - \sqrt{8})

Option 1 =

\sqrt{7} - 2\sqrt{2} = \sqrt{7} - \sqrt{8}

, which is satisfied
Option 3 =

\sqrt{8} - \sqrt{7} = - (\sqrt{7} - \sqrt{8})

, which is also satisfied

Answer: (4) More than one of the above

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