Real Numbers

\(\sqrt{36}=6\) and \(\sqrt{121}=11\). Therefore, \(\sqrt{36}-\sqrt{121}=6-11=-5\). The absolute value of a number is never negative, so \(\left|-5\right|=5\). Option C is a close distractor because it is the value before applying the absolute value. Exam tip: evaluate the square roots first, then take the absolute value of the difference.

In real numbers, \(\sqrt{a^2}=|a|\), because the principal square root is always non-negative. Therefore, \(\sqrt{(-12)^2}=|-12|=12\). The option \(-12\) is incorrect: although its square is 144, the principal square root of 144 is 12. Exam tip: Do not simplify \(\sqrt{x^2}\) directly as \(x\); write it as \(|x|\).

Use \(\sqrt{x^2}=|x|\), not simply \(x\) in every case. Since \(x=-15\), \(\sqrt{x^2}=|-15|=15\). Therefore, \(\sqrt{x^2}-x=15-(-15)=30\). The option \(0\) results from the incorrect assumption that \(\sqrt{x^2}=x\) for a negative value of \(x\). Exam tip: the principal square root is always non-negative, so write \(\sqrt{x^2}=|x|\).

Multiplying by the conjugate (6-\sqrt{35}) makes the denominator (36-35=1). So the rationalised form is (6-\sqrt{35}).

( \sqrt{500}=10\sqrt{5} ), ( \sqrt{180}=6\sqrt{5} ), and ( \sqrt{45}=3\sqrt{5} ). Therefore (10\sqrt{5}-6\sqrt{5}+3\sqrt{5}=7\sqrt{5}).

(2\sqrt{147}=14\sqrt{3}), (3\sqrt{75}=15\sqrt{3}), and ( \sqrt{300}=10\sqrt{3} ). The coefficient is (14-15+10=9), so the value is (9\sqrt{3}).

(4\sqrt{72}=24\sqrt{2}), (2\sqrt{200}=20\sqrt{2}), and (5\sqrt{50}=25\sqrt{2}). Therefore the result should be (19\sqrt{2}).

(2\sqrt{8}=4\sqrt{2}) and ( \sqrt{50}=5\sqrt{2} ), so the bracket is ( -\sqrt{2} ). ( \sqrt{18}\times(-\sqrt{2})=-\sqrt{36}=-6 ).

The fraction \(\frac{77}{480}\) is already in lowest terms, and \(480=2^5\times3\times5\). A rational number has a terminating decimal expansion only when the denominator in lowest form has no prime factors other than 2 and 5. Since 3 is also present, its decimal expansion is non-terminating recurring. Option C is incorrect because non-terminating non-recurring decimals represent irrational numbers. Exam tip: First reduce the fraction, then check the prime factors of its denominator.

( (5\sqrt{3})^2=75 ) and ( (2\sqrt{5})^2=20 ). The middle term (2\times5\sqrt{3}\times2\sqrt{5}=20\sqrt{15}) is subtracted.

\(\sqrt{2}+\sqrt{3}\) is irrational. If it were equal to a rational number \(r\), then on squaring we would get \(r^2=5+2\sqrt{6}\). This would make \(\sqrt{6}\) rational, which is impossible; hence the sum is irrational. In option B, \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), so it is rational. Exam tip: To test a sum of square roots, assume it is rational and square both sides.

Multiplying by the conjugate gives denominator (18-20=-2). So the form can be written as ( \frac{2\sqrt{5}-3\sqrt{2}}{2} ).

Multiplying by the conjugate ( \sqrt{7}+\sqrt{5} ) gives denominator (7-5=2). The numerator also has (2), so the answer is ( \sqrt{7}+\sqrt{5} ).

Use the conjugate-pair identity with a common denominator. Let a=3+√2 and b=3−√2. Then a/b+b/a=(a²+b²)/(ab). The denominator is ab=(3+√2)(3−√2)=3²−(√2)²=9−2=7. For the numerator, expand a²+b²: (3+√2)²+(3−√2)²=(9+6√2+2)+(9−6√2+2)=22. The irrational cross-terms cancel because they have opposite signs. Hence the original value is 22/7, so option A is correct. Option B is half of the correct numerator, usually caused by dropping one squared term. Option C results from an incomplete or incorrect expansion. Option D ignores the denominator and the actual values of the conjugate expressions, so it cannot be correct. The denominator is nonzero, so the operations are valid.

Use the identity \((a+b)^2-(a-b)^2=4ab\). Here, \(a=\sqrt{5}\) and \(b=\sqrt{3}\). Therefore, the value is \(4\times\sqrt{5}\times\sqrt{3}=4\sqrt{15}\). The option 16 may result from ignoring the effect of the middle terms in the two squares. Exam tip: For expressions of this form, applying the identity is quicker and safer than expanding both squares separately.

( \sqrt{12}=2\sqrt{3} ) and ( \sqrt{27}=3\sqrt{3} ). Dividing (5\sqrt{3}) by ( \sqrt{3} ) gives (5).

Adding a rational number to an irrational number always gives an irrational number. Hence \(x+5\) is irrational because 5 is rational. Option B is not always true: if \(x=\sqrt{2}\), then \(x^2=2\), which is rational. In options C and D, \(x\neq0\), so the value is 1, a rational number. Exam tip: irrational ± rational is always irrational.

The correct answer is \(\frac{21}{84}\), because the fraction must first be reduced to lowest terms: \(\frac{21}{84}=\frac{1}{4}=0.25\). Its denominator in lowest form is \(4=2^2\), so its decimal expansion terminates. The student's error is checking prime factors before cancelling common factors. In contrast, \(\frac{7}{12}\) has a factor of \(3\) in its denominator in lowest terms, so its decimal expansion is non-terminating recurring. Exam tip: Always reduce a fraction to lowest terms before applying the terminating-decimal rule.

In \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing, so no fixed block repeats. Its decimal expansion is non-terminating and non-repeating; therefore, it is irrational. Although \(0.272727\ldots\) is non-terminating, the block 27 repeats, so it is rational. Also, \(0.125\) terminates and \(\sqrt{121}=11\) is an integer. Exam tip: A non-terminating decimal is irrational only when it has no repeating pattern.

( \frac{11}{5}=2.2 ), ( \sqrt{5}\approx2.236 ), and then (2.3). So the ascending order is ( \frac{11}{5}<\sqrt{5}<2.3 ).