Year 9 Maths Core Pythagoras Theorem

Squares, Square Roots and Surds

10 practice questions 2 video lessons Theory + worked examples

Theory

Squares, Square Roots and Surds — Theory

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Theory — Squares, Square Roots and Surds

          Key concept: A square root of a number n is a value that, when multiplied by itself, gives n. If the square root simplifies to a whole number or fraction, it is rational. If it does not simplify exactly, it is left in surd form (e.g. 34), which is an irrational number.
        

To evaluate expressions involving squares and square roots, first compute the squares, then simplify under the radical sign. If the result is a perfect square the answer is a whole number; otherwise it is expressed as a surd or rounded to a given number of decimal places.

Recall the key identity: $\sqrt{a^{2}} = a$ for $a \geq 0$ . Also note that $\sqrt{a^{2} + b^{2}} \neq a + b$ in general — you must evaluate the expression inside the square root first.

Example 1

Evaluate the following

\sqrt{4^{2} + 3^{2}}

ii)

\sqrt{17^{2} - 15^{2}}

Solution

\begin{aligned} \sqrt{4^{2} + 3^{2}} & = \sqrt{16 + 9} \\ = \sqrt{25} \\ = \sqrt{5 \times 5} \\ = 5 \end{aligned}

ii)

\begin{aligned} \sqrt{17^{2} - 15^{2}} & = \sqrt{289 - 225} \\ = \sqrt{64} \\ = \sqrt{8 \times 8} \\ = 8 \end{aligned}

Example 2

Evaluate the following

\sqrt{5^{2} + 3^{2}}

ii)

\sqrt{5^{2} - 2^{2}}

Solution

\begin{aligned} \sqrt{5^{2} + 3^{2}} & = \sqrt{25 + 9} \\ = \sqrt{34} & (as a surd) \\ = 5.8 & (one dp) \end{aligned}

ii)

\begin{aligned} \sqrt{5^{2} - 2^{2}} & = \sqrt{25 - 4} \\ = \sqrt{21} & (as a surd) \\ = 4.6 & (one dp) \end{aligned}

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Pythagoras Theorem

Squares, Square Roots and Surds

📖 Theory

Theory — Squares, Square Roots and Surds

🎬 Video Lessons

✏️ Practice Questions

Theory

Video Lessons

Practice Questions