CBSE Class 9 Mathematics Number System Worksheet Set 04

Question 1. Visualize 3.756 on the number line, using successive magnification
Answer: To locate \( 3.756 \) on the number line using the method of successive magnification, follow these steps:
1. The number \( 3.756 \) lies between the integers \( 3 \) and \( 4 \). Divide the unit segment between \( 3 \) and \( 4 \) into 10 equal intervals of length \( 0.1 \). The number lies in the sub-interval between \( 3.7 \) and \( 3.8 \).
2. Magnify the segment from \( 3.7 \) to \( 3.8 \) and partition it into 10 equal divisions of length \( 0.01 \). The value \( 3.756 \) lies between \( 3.75 \) and \( 3.76 \).
3. Magnify the segment between \( 3.75 \) and \( 3.76 \) and divide it into 10 equal sub-divisions of length \( 0.001 \).
4. Counting 6 subdivisions to the right of \( 3.75 \) identifies the exact position of the real number \( 3.756 \).
In simple words: Zoom into the number line step-by-step. First look between 3 and 4, then between 3.7 and 3.8, and finally between 3.75 and 3.76 to spot 3.756 at the sixth tick mark.

Exam Tip: Always show each magnification step on a separate horizontal line with clearly labeled endpoints and subdivisions to earn full marks.

Question 2. Represent √3.5 on the number line
Answer: To construct and represent \( \sqrt{3.5} \) geometrically on the number line:
1. Mark a fixed point \( A \) and draw a straight horizontal line segment \( AB = 3.5\text{ units} \).
2. Extend line segment \( AB \) past point \( B \) to a point \( C \) such that \( BC = 1\text{ unit} \). The total length of segment \( AC \) becomes \( 3.5 + 1 = 4.5\text{ units} \).
3. Determine the midpoint \( O \) of segment \( AC \) using a compass, such that \( OA = OC = \frac{4.5}{2} = 2.25\text{ units} \).
4. With point \( O \) as the center and radius \( OA = 2.25\text{ units} \), draw a semicircle passing through points \( A \) and \( C \).
5. Construct a perpendicular line to \( AC \) passing through point \( B \), intersecting the arc of the semicircle at point \( D \). By the geometric mean theorem, the length of segment \( BD = \sqrt{AB \times BC} = \sqrt{3.5 \times 1} = \sqrt{3.5}\text{ units} \).
6. Treating point \( B \) as the origin (\( 0 \)) on the number line, place the compass needle at \( B \) with radius \( BD \), and draw an arc cutting the extended line to the right at point \( P \). The coordinate of point \( P \) represents \( \sqrt{3.5} \).
In simple words: Draw a line of 3.5 units, add 1 extra unit, and draw a semicircle over the whole segment. The vertical height drawn at the 3.5 mark gives the exact length of square root of 3.5, which is then rotated onto the line.

Exam Tip: Remember to specify point B as origin 0 when drawing the arc onto the number line; measuring from point O instead of point B is a common pitfall.

Question 3. If x = 3+2√2, find the value of x² + 1/x²
Answer: Given the value of \( x = 3 + 2\sqrt{2} \), determine its reciprocal by rationalizing the denominator:
\( \frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \)
\( = \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2} \)
Summing \( x \) and \( \frac{1}{x} \):
\( x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \)
Squaring both sides of this sum yields:
\( \left(x + \frac{1}{x}\right)^2 = 6^2 \)
\( x^2 + \frac{1}{x^2} + 2 = 36 \)
Subtract 2 from both sides:
\( \implies x^2 + \frac{1}{x^2} = 36 - 2 = 34 \)
In simple words: Find the reciprocal of x by multiplying numerator and denominator by 3 minus 2 root 2. Adding x and 1/x gives 6, and squaring 6 minus 2 produces 34.

Exam Tip: Use the standard algebraic identity \( x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \) to directly calculate the answer without squaring the large irrational binomials separately.

Question 4. If x = 2 + √5, Prove that x² + 1/x² = 18
Answer: Given \( x = \sqrt{5} + 2 \), evaluate the reciprocal \( \frac{1}{x} \) by multiplying the numerator and denominator by its conjugate:
\( \frac{1}{x} = \frac{1}{\sqrt{5} + 2} \times \frac{\sqrt{5} - 2}{\sqrt{5} - 2} \)
\( = \frac{\sqrt{5} - 2}{(\sqrt{5})^2 - 2^2} = \frac{\sqrt{5} - 2}{5 - 4} = \sqrt{5} - 2 \)
Compute the sum of \( x \) and \( \frac{1}{x} \):
\( x + \frac{1}{x} = (\sqrt{5} + 2) + (\sqrt{5} - 2) = 2\sqrt{5} \)
Now compute the expression on the left-hand side:
\( \text{L.H.S.} = x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \)
Substitute \( x + \frac{1}{x} = 2\sqrt{5} \):
\( = (2\sqrt{5})^2 - 2 = 4(5) - 2 = 20 - 2 = 18 = \text{R.H.S.} \)
Hence proved.
In simple words: Flip x and rationalize it to get root 5 minus 2. Their sum is 2 root 5, and squaring that sum gives 20, which minus 2 equals 18.

Exam Tip: Writing \( 2 + \sqrt{5} \) as \( \sqrt{5} + 2 \) keeps the denominator positive when applying \( a^2 - b^2 \), preventing common negative sign errors.

Question 5. Express 1. 3 2 + 0.35 as a fraction in simplest form.
Answer: Let the repeating decimal terms be converted into fractional form separately:
First, consider \( x = 1.\overline{32} = 1.323232... \)
Multiplying by 100 shifts the repetend:
\( 100x = 132.3232... \)
Subtracting the original equation:
\( 100x - x = 132.3232... - 1.3232... \)
\( \implies 99x = 131 \implies x = \frac{131}{99} \)
Next, consider \( y = 0.\overline{35} = 0.353535... \)
Multiplying by 100 gives:
\( 100y = 35.3535... \)
Subtracting the original equation:
\( 100y - y = 35.3535... - 0.3535... \)
\( \implies 99y = 35 \implies y = \frac{35}{99} \)
Adding the two fractions:
\( x + y = \frac{131}{99} + \frac{35}{99} = \frac{131 + 35}{99} = \frac{166}{99} \)
Because 166 and 99 share no common factors other than 1, the fraction is in its simplest form.
In simple words: Turn each recurring decimal into a fraction over 99. Adding 131 over 99 and 35 over 99 results in 166 over 99.

Exam Tip: Do not add non-terminating recurring decimals directly as decimals; always convert them to \( \frac{p}{q} \) form first to avoid incorrect rounding.

Question 6. Rationalise the denominator 1 / (√6 + √5 - √11)
Answer: Group the first two terms in the denominator:
\( \frac{1}{(\sqrt{6} + \sqrt{5}) - \sqrt{11}} \)
Multiply the numerator and denominator by the conjugate factor \( (\sqrt{6} + \sqrt{5}) + \sqrt{11} \):
\( = \frac{(\sqrt{6} + \sqrt{5}) + \sqrt{11}}{[(\sqrt{6} + \sqrt{5}) - \sqrt{11}][(\sqrt{6} + \sqrt{5}) + \sqrt{11}]} \)
Apply the difference of squares identity in the denominator:
Denominator \( = (\sqrt{6} + \sqrt{5})^2 - (\sqrt{11})^2 \)
\( = (6 + 5 + 2\sqrt{30}) - 11 = 11 + 2\sqrt{30} - 11 = 2\sqrt{30} \)
This simplifies the overall expression to:
\( = \frac{\sqrt{6} + \sqrt{5} + \sqrt{11}}{2\sqrt{30}} \)
Rationalize the remaining radical by multiplying numerator and denominator by \( \sqrt{30} \):
\( = \frac{\sqrt{30}(\sqrt{6} + \sqrt{5} + \sqrt{11})}{2 \times 30} \)
\( = \frac{\sqrt{180} + \sqrt{150} + \sqrt{330}}{60} \)
Simplifying each individual square root:
\( \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \)
\( \sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6} \)
Thus, the final rationalized expression is:
\( = \frac{6\sqrt{5} + 5\sqrt{6} + \sqrt{330}}{60} \)
In simple words: Group root 6 and root 5 together as one unit and multiply by its conjugate. The 11s cancel out in the denominator, leaving only root 30, which is easily rationalized in the second step.

Exam Tip: Notice that \( (\sqrt{6})^2 + (\sqrt{5})^2 = 11 \); grouping \( (\sqrt{6} + \sqrt{5}) \) together ensures the integer parts cancel completely in the first step.

Question 7. If a and b are rational numbers, find a and b
(a) \( \frac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} = a + b\sqrt{6} \)
(b) \( \frac{\sqrt{5} - 2}{\sqrt{5} + 2} - \frac{\sqrt{5} + 2}{\sqrt{5} - 2} = a + b\sqrt{5} \)
Answer:
(a) Multiply the numerator and denominator of the left-hand side by the conjugate \( 3\sqrt{2} + 2\sqrt{3} \):
\( \text{L.H.S.} = \frac{(\sqrt{2} + \sqrt{3})(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} - 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3})} \)
Denominator \( = (3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6 \)
Numerator \( = 3(2) + 2\sqrt{6} + 3\sqrt{6} + 2(3) = 6 + 5\sqrt{6} + 6 = 12 + 5\sqrt{6} \)
\( = \frac{12 + 5\sqrt{6}}{6} = \frac{12}{6} + \frac{5}{6}\sqrt{6} = 2 + \frac{5}{6}\sqrt{6} \)
Equating with \( a + b\sqrt{6} \) gives:
\( a = 2, \quad b = \frac{5}{6} \)

(b) Simplify the left-hand side by taking the common denominator:
\( \text{L.H.S.} = \frac{(\sqrt{5} - 2)^2 - (\sqrt{5} + 2)^2}{(\sqrt{5} + 2)(\sqrt{5} - 2)} \)
Denominator \( = (\sqrt{5})^2 - 2^2 = 5 - 4 = 1 \)
Numerator \( = (5 + 4 - 4\sqrt{5}) - (5 + 4 + 4\sqrt{5}) = (9 - 4\sqrt{5}) - (9 + 4\sqrt{5}) = -8\sqrt{5} \)
\( = \frac{-8\sqrt{5}}{1} = 0 - 8\sqrt{5} \)
Equating with \( a + b\sqrt{5} \) gives:
\( a = 0, \quad b = -8 \)
In simple words: Rationalize the denominators on the left side so that the terms match the pattern of a rational number plus a radical term, then compare coefficients to find a and b.

Exam Tip: Clearly separate the rational term and the coefficient of the irrational term; when a pure root term remains like \( -8\sqrt{5} \), state explicitly that \( a = 0 \).

Question 8. Simplify: a) 1/(1 + √2) + 1/(√2 + √3) + 1/(√3 + √4)
Answer: Rationalize each fraction individually by multiplying by its conjugate:
First term:
\( \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1 \)
Second term:
\( \frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2} \)
Third term:
\( \frac{1}{\sqrt{4} + \sqrt{3}} \times \frac{\sqrt{4} - \sqrt{3}}{\sqrt{4} - \sqrt{3}} = \frac{\sqrt{4} - \sqrt{3}}{4 - 3} = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3} \)
Sum all three rationalized expressions:
\( (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) \)
All intermediate radical terms cancel:
\( = \sqrt{4} - 1 = 2 - 1 = 1 \)
In simple words: When each fraction is rationalized, the denominators become 1 and the middle square roots cancel out in a chain, leaving only 2 minus 1, which equals 1.

Exam Tip: Notice the telescoping sum pattern where consecutive terms cancel out; write the larger radical first in each denominator to avoid negative signs.

Question 9. If a = (√3+√2)/(√3 - √2) and b = (√3 – √2)/(√3 + √2), find the value of a² + b²
Answer: First, rationalize both \( a \) and \( b \):
\( a = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} + \sqrt{2})^2}{3 - 2} = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6} \)
\( b = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{(\sqrt{3} - \sqrt{2})^2}{3 - 2} = 3 + 2 - 2\sqrt{6} = 5 - 2\sqrt{6} \)
Calculate their sum and product:
\( a + b = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10 \)
\( ab = (5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 5^2 - (2\sqrt{6})^2 = 25 - 24 = 1 \)
Using the identity \( a^2 + b^2 = (a + b)^2 - 2ab \):
\( a^2 + b^2 = (10)^2 - 2(1) = 100 - 2 = 98 \)
In simple words: Rationalizing gives 5 plus 2 root 6 for a and 5 minus 2 root 6 for b. Their sum is 10 and product is 1, so squaring 10 and subtracting 2 gives 98.

Exam Tip: Always calculate \( a + b \) and \( ab \) first; applying \( (a + b)^2 - 2ab \) is much faster than squaring each binomial individually.

Question 10. If a = 9 - 4√5, find the value of [ a – 1/a ]²
Answer: Given \( a = 9 - 4\sqrt{5} \), determine the reciprocal \( \frac{1}{a} \):
\( \frac{1}{a} = \frac{1}{9 - 4\sqrt{5}} \times \frac{9 + 4\sqrt{5}}{9 + 4\sqrt{5}} = \frac{9 + 4\sqrt{5}}{9^2 - (4\sqrt{5})^2} = \frac{9 + 4\sqrt{5}}{81 - 80} = 9 + 4\sqrt{5} \)
Find the difference between \( a \) and \( \frac{1}{a} \):
\( a - \frac{1}{a} = (9 - 4\sqrt{5}) - (9 + 4\sqrt{5}) = 9 - 4\sqrt{5} - 9 - 4\sqrt{5} = -8\sqrt{5} \)
Now square this difference:
\( \left[a - \frac{1}{a}\right]^2 = (-8\sqrt{5})^2 = (-8)^2 \times (\sqrt{5})^2 = 64 \times 5 = 320 \)
In simple words: Rationalizing 1/a gives 9 plus 4 root 5. Subtracting it from a leaves minus 8 root 5, and squaring that value results in 320.

Exam Tip: Distribute the negative sign carefully across both terms of \( \frac{1}{a} \) during subtraction so that the integer terms cancel and the root terms combine correctly.

Question 11. If x = 1 - √2, find the value of [ x – 1/x ]³
Answer: Given \( x = 1 - \sqrt{2} \), evaluate its reciprocal:
\( \frac{1}{x} = \frac{1}{1 - \sqrt{2}} \times \frac{1 + \sqrt{2}}{1 + \sqrt{2}} = \frac{1 + \sqrt{2}}{1^2 - (\sqrt{2})^2} = \frac{1 + \sqrt{2}}{1 - 2} = \frac{1 + \sqrt{2}}{-1} = -1 - \sqrt{2} \)
Compute the difference \( x - \frac{1}{x} \):
\( x - \frac{1}{x} = (1 - \sqrt{2}) - (-1 - \sqrt{2}) = 1 - \sqrt{2} + 1 + \sqrt{2} = 2 \)
Now cube the resulting value:
\( \left[x - \frac{1}{x}\right]^3 = (2)^3 = 8 \)
In simple words: The reciprocal of 1 minus root 2 simplifies to minus 1 minus root 2. Subtracting the two gives exactly 2, and 2 cubed is 8.

Exam Tip: Do not overlook the negative sign in the denominator \( 1 - 2 = -1 \) when rationalizing; multiplying through by \( -1 \) is crucial for the correct value.

Question 12. If x = 3 + 2 √2, find the value of [ √x - 1/√x ]
Answer: Given \( x = 3 + 2\sqrt{2} \), express \( x \) as a perfect square:
\( x = 2 + 1 + 2\sqrt{2} = (\sqrt{2})^2 + 1^2 + 2(\sqrt{2})(1) = (\sqrt{2} + 1)^2 \)
Taking the positive square root:
\( \sqrt{x} = \sqrt{2} + 1 \)
Now compute the reciprocal \( \frac{1}{\sqrt{x}} \):
\( \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1 \)
Subtract the two quantities:
\( \left[\sqrt{x} - \frac{1}{\sqrt{x}}\right] = (\sqrt{2} + 1) - (\sqrt{2} - 1) = \sqrt{2} + 1 - \sqrt{2} + 1 = 2 \)
(Note: If the expression is cubed, \( \left[\sqrt{x} - \frac{1}{\sqrt{x}}\right]^3 = 2^3 = 8 \)).
In simple words: Rewrite x as root 2 plus 1 squared so that root x is simply root 2 plus 1. Subtracting its reciprocal root 2 minus 1 leaves 2.

Exam Tip: When given \( x = a + 2\sqrt{b} \), split the integer into two parts whose sum is \( a \) and whose product is \( b \) to write \( x \) as a perfect square.

Question 13. If x = 0.125, find the value of (1/x)¹/³
Answer: Convert the decimal into fractional form:
\( x = 0.125 = \frac{125}{1000} = \frac{1}{8} \)
Take the reciprocal of \( x \):
\( \frac{1}{x} = 8 \)
Now evaluate the cube root:
\( \left(\frac{1}{x}\right)^{1/3} = 8^{1/3} = (2^3)^{1/3} = 2^{3 \times \frac{1}{3}} = 2^1 = 2 \)
In simple words: 0.125 is the fraction 1 over 8. Its reciprocal is 8, and the cube root of 8 is 2.

Exam Tip: Expressing decimals as powers of prime fractions makes calculating fractional exponents immediate and error-free.

Question 14. If x = √3 + √2 / (√3 - √2) then find the value of x²
Answer: First, rationalize the denominator of \( x \):
\( x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \)
\( = \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{3 + 2 + 2\sqrt{6}}{3 - 2} = 5 + 2\sqrt{6} \)
Now square the resulting simplified binomial:
\( x^2 = (5 + 2\sqrt{6})^2 \)
\( = 5^2 + (2\sqrt{6})^2 + 2(5)(2\sqrt{6}) \)
\( = 25 + 24 + 20\sqrt{6} = 49 + 20\sqrt{6} \)
In simple words: Rationalizing x produces 5 plus 2 root 6. Squaring this using the formula a squared plus b squared plus 2ab yields 49 plus 20 root 6.

Exam Tip: Square the whole rationalized expression using \( (a + b)^2 = a^2 + 2ab + b^2 \); remember that \( (2\sqrt{6})^2 = 4 \times 6 = 24 \).

Question 15. If x = 1 / (2 - √3), find the value of x³ – 2x² – 7x + 5
Answer: Rationalize the given value of \( x \):
\( x = \frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3} \)
Rearrange and square to eliminate the square root:
\( x - 2 = \sqrt{3} \)
\( (x - 2)^2 = (\sqrt{3})^2 \)
\( x^2 - 4x + 4 = 3 \)
\( \implies x^2 - 4x + 1 = 0 \)
Express the given cubic polynomial in terms of \( x^2 - 4x + 1 \):
\( x^3 - 2x^2 - 7x + 5 = x(x^2 - 4x + 1) + 2x^2 - 8x + 5 \)
\( = x(x^2 - 4x + 1) + 2(x^2 - 4x + 1) + 3 \)
\( = (x + 2)(x^2 - 4x + 1) + 3 \)
Since \( x^2 - 4x + 1 = 0 \):
\( = (x + 2)(0) + 3 = 3 \)
In simple words: Rationalize x to 2 plus root 3, then build the quadratic equation x squared minus 4x plus 1 equals zero. Dividing the big polynomial by this quadratic leaves a remainder of 3.

Exam Tip: Form the zero-value quadratic polynomial \( x^2 - 4x + 1 = 0 \) first rather than substituting the radical into every power of \( x \); polynomial division saves time and prevents calculation mistakes.

Question 16. Find four rational numbers between 3/5 and 4/5
Answer: To find 4 rational numbers between \( \frac{3}{5} \) and \( \frac{4}{5} \), multiply the numerator and denominator of both numbers by \( 4 + 1 = 5 \):
\( \frac{3}{5} = \frac{3 \times 5}{5 \times 5} = \frac{15}{25} \)
\( \frac{4}{5} = \frac{4 \times 5}{5 \times 5} = \frac{20}{25} \)
Selecting integers between 15 and 20 for the numerator, four rational numbers are:
\( \frac{16}{25}, \frac{17}{25}, \frac{18}{25}, \text{ and } \frac{19}{25} \)
In simple words: Scale up both fractions by multiplying top and bottom by 5 to create space between the numerators, then pick the fractions in between: 16/25, 17/25, 18/25, and 19/25.

Exam Tip: To find \( n \) rational numbers between two fractions with the same denominator, multiply the numerator and denominator by \( n + 1 \) to create enough integer steps between them.

Question 17. Find two irrational numbers lying between √2 and √3
Answer: The approximate decimal expansions of the given roots are:
\( \sqrt{2} \approx 1.4142... \)
\( \sqrt{3} \approx 1.7320... \)
Any non-terminating and non-recurring decimal falling between these two limits is an irrational number. Two such numbers are:
- \( 1.5050050005... \)
- \( 1.6060060006... \)
In simple words: Pick numbers between 1.414 and 1.732 and write them with an ever-expanding pattern of zeros so that they never end and never repeat.

Exam Tip: Using a clear, expanding non-repeating pattern (like 1.5050050005...) proves to the examiner that the decimal is non-terminating and non-recurring.

Question 18. Find two rational and irrational numbers between 0.3101 and 0.3222
Answer: The specified interval is between \( 0.3101 \) and \( 0.3222 \).
(i) Two rational numbers in this range (terminating decimals):
- \( 0.312 = \frac{312}{1000} = \frac{39}{125} \)
- \( 0.320 = \frac{32}{100} = \frac{8}{25} \)
(ii) Two irrational numbers in this range (non-terminating, non-recurring decimals):
- \( 0.311010010001... \)
- \( 0.321010010001... \)
In simple words: Simple terminating decimals like 0.312 and 0.320 are rational, while numbers with non-repeating infinite patterns like 0.31101001... are irrational.

Exam Tip: State the defining characteristics: terminating decimals represent rational numbers, while non-terminating non-repeating decimals represent irrational numbers.

Question 19. Simplify the following:
(a) \( \left(\frac{576}{625}\right)^{-1/2} \)
(b) \( \left(\frac{343}{1000}\right)^{-1/3} \)
(c) \( \left(-\frac{1}{27}\right)^{-2/3} \)
(d) \( (0.008)^{4/3} \)
(e) \( (729)^{-1/6} \)
Answer:
(a) Apply the negative exponent rule to invert the base:
\( \left(\frac{576}{625}\right)^{-1/2} = \left(\frac{625}{576}\right)^{1/2} = \sqrt{\frac{25^2}{24^2}} = \frac{25}{24} \)

(b) Invert the base and express in prime powers:
\( \left(\frac{343}{1000}\right)^{-1/3} = \left(\frac{1000}{343}\right)^{1/3} = \left(\frac{10^3}{7^3}\right)^{1/3} = \frac{10}{7} \)

(c) Invert the fraction and rewrite \( -27 \) as a cube:
\( \left(-\frac{1}{27}\right)^{-2/3} = (-27)^{2/3} = [(-3)^3]^{2/3} = (-3)^2 = 9 \)

(d) Convert the decimal to a fraction:
\( (0.008)^{4/3} = \left(\frac{8}{1000}\right)^{4/3} = \left[\left(\frac{2}{10}\right)^3\right]^{4/3} = \left(\frac{2}{10}\right)^4 = (0.2)^4 = 0.0016 = \frac{1}{625} \)

(e) Express 729 as a power of 3:
\( (729)^{-1/6} = (3^6)^{-1/6} = 3^{6 \times \left(-\frac{1}{6}\right)} = 3^{-1} = \frac{1}{3} \)
In simple words: Negative powers flip the fraction upside down. Rewrite each base as a power that matches the root in the exponent so they cancel cleanly.

Exam Tip: First use \( \left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m \) to turn negative exponents positive before simplifying powers.

Question 20. Simplify and express the result in the simplest form: ((25)³/² x (243)²/⁵) / ((16)⁵/⁴ x (8)⁴/³)
Answer: Express each base in terms of its prime factors:
- For the numerator:
\( (25)^{3/2} = (5^2)^{3/2} = 5^{2 \times \frac{3}{2}} = 5^3 = 125 \)
\( (243)^{2/5} = (3^5)^{2/5} = 3^{5 \times \frac{2}{5}} = 3^2 = 9 \)
Numerator \( = 125 \times 9 = 1125 \)
- For the denominator:
\( (16)^{5/4} = (2^4)^{5/4} = 2^{4 \times \frac{5}{4}} = 2^5 = 32 \)
\( (8)^{4/3} = (2^3)^{4/3} = 2^{3 \times \frac{4}{3}} = 2^4 = 16 \)
Denominator \( = 32 \times 16 = 512 \)
Dividing numerator by denominator gives:
\( \frac{1125}{512} \)
In simple words: Change every number into prime powers so the fractional powers cancel out. The top becomes 125 times 9, and the bottom becomes 32 times 16, giving 1125 over 512.

Exam Tip: Convert every base into prime base form (\( 25 = 5^2 \), \( 243 = 3^5 \), \( 16 = 2^4 \), \( 8 = 2^3 \)) to cancel fractional exponents directly.

Question 21. Find the value x, if 5^(x - 3) x 3^(2x – 8) = 225
Answer: Find the prime factorization of 225:
\( 225 = 25 \times 9 = 5^2 \times 3^2 \)
Equate the given expression to this prime factorization:
\( 5^{x - 3} \times 3^{2x - 8} = 5^2 \times 3^2 \)
Comparing exponents of the identical prime base 5:
\( x - 3 = 2 \)
\( \implies x = 2 + 3 = 5 \)
Verifying with the exponents of prime base 3:
\( 2x - 8 = 2 \)
\( \implies 2x = 10 \implies x = 5 \)
Both bases consistently give \( x = 5 \).
In simple words: Break 225 into 5 squared times 3 squared. Match the powers of 5 and 3 to see that x minus 3 equals 2, meaning x is 5.

Exam Tip: Always verify that both prime base equations yield the same value of \( x \); this confirms your solution is completely correct.

Question 22. Solve:
(a) \( 49 \times 7^x = (343)^{1/3} \)
(b) \( 2^x = (128)^{1/7} \times (\sqrt{2})^4 \)
(c) If \( 3^x = \frac{9}{27^x} \), find x
(d) \( \left(\frac{1}{7}\right)^{4 - 2x} = \sqrt{7} \)
Answer:
(a) Express all terms with base 7:
\( 7^2 \times 7^x = (7^3)^{1/3} \)
\( 7^{x + 2} = 7^1 \)
Equating exponents:
\( \implies x + 2 = 1 \implies x = -1 \)

(b) Express all terms with base 2:
\( 128 = 2^7 \implies (128)^{1/7} = (2^7)^{1/7} = 2^1 = 2 \)
\( (\sqrt{2})^4 = (2^{1/2})^4 = 2^2 = 4 \)
\( 2^x = 2 \times 4 = 8 = 2^3 \)
Equating exponents:
\( \implies x = 3 \)

(c) Multiply both sides by \( 27^x \):
\( 3^x \times 27^x = 9 \)
Express using base 3:
\( 3^x \times (3^3)^x = 3^2 \)
\( 3^x \times 3^{3x} = 3^2 \)
\( 3^{4x} = 3^2 \)
Equating exponents:
\( \implies 4x = 2 \implies x = \frac{2}{4} = \frac{1}{2} \)

(d) Express both sides with base 7:
\( (7^{-1})^{4 - 2x} = 7^{1/2} \)
\( 7^{-(4 - 2x)} = 7^{1/2} \)
\( 7^{2x - 4} = 7^{1/2} \)
Equating exponents:
\( \implies 2x - 4 = \frac{1}{2} \)
\( \implies 2x = 4 + \frac{1}{2} = \frac{9}{2} \)
\( \implies x = \frac{9}{4} \)
In simple words: Convert every part of each equation to have the exact same base number. Once the bases match, set the exponents equal to solve for x.

Exam Tip: Watch out for negative exponents when rewriting fractions like \( \frac{1}{7} \) as \( 7^{-1} \); distribute the negative sign through the entire power expression \( -(4 - 2x) = 2x - 4 \).

Question 23. Evaluate:
(a) \( 125^{-1/3} \times 27^{1/3} (6^2 + 8^2)^{1/2} \)
(b) \( (17^2 - 8^2)^{1/2} \)
(c) \( 64^{1/3} (64^{1/3} - 64^{2/3}) \)
Answer:
(a) Evaluate each factor:
\( 125^{-1/3} = (5^3)^{-1/3} = 5^{-1} = \frac{1}{5} \)
\( 27^{1/3} = (3^3)^{1/3} = 3^1 = 3 \)
\( (6^2 + 8^2)^{1/2} = (36 + 64)^{1/2} = (100)^{1/2} = 10 \)
Multiplying the three factors together:
\( = \frac{1}{5} \times 3 \times 10 = 3 \times 2 = 6 \)

(b) Use the difference of squares identity:
\( 17^2 - 8^2 = (17 - 8)(17 + 8) = 9 \times 25 = 225 \)
Taking the square root:
\( (225)^{1/2} = \sqrt{225} = 15 \)

(c) Express 64 as a power of 4:
\( 64^{1/3} = (4^3)^{1/3} = 4 \)
\( 64^{2/3} = (4^3)^{2/3} = 4^2 = 16 \)
Substitute into the expression:
\( 4(4 - 16) = 4(-12) = -48 \)
In simple words: Simplify roots and powers step-by-step. In (a), 1/5 times 3 times 10 gives 6. In (b), 289 minus 64 is 225 whose root is 15. In (c), 4 times minus 12 is minus 48.

Exam Tip: Never distribute a square root over addition or subtraction; always calculate \( 6^2 + 8^2 = 100 \) inside the parentheses before taking the square root.

Question 24. Simplify:
(a) \( \sqrt{45} + \sqrt{80} - 3\sqrt{20} \)
(b) \( 7\sqrt{6} - \sqrt{252} - \sqrt{294} + 6\sqrt{7} \)
(c) \( 4\sqrt{28} + 3\sqrt{7} \)
Answer:
(a) Express each radical in simplest form:
\( \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \)
\( \sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5} \)
\( 3\sqrt{20} = 3\sqrt{4 \times 5} = 3(2\sqrt{5}) = 6\sqrt{5} \)
Combining like radicals:
\( = 3\sqrt{5} + 4\sqrt{5} - 6\sqrt{5} = (3 + 4 - 6)\sqrt{5} = \sqrt{5} \)

(b) Extract perfect squares from each radical:
\( \sqrt{252} = \sqrt{36 \times 7} = 6\sqrt{7} \)
\( \sqrt{294} = \sqrt{49 \times 6} = 7\sqrt{6} \)
Substitute back into the expression:
\( = 7\sqrt{6} - 6\sqrt{7} - 7\sqrt{6} + 6\sqrt{7} \)
Group like terms:
\( = (7\sqrt{6} - 7\sqrt{6}) + (6\sqrt{7} - 6\sqrt{7}) = 0 + 0 = 0 \)

(c) Simplify the first square root:
\( \sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7} \)
\( 4\sqrt{28} = 4(2\sqrt{7}) = 8\sqrt{7} \)
Adding the like radicals:
\( = 8\sqrt{7} + 3\sqrt{7} = (8 + 3)\sqrt{7} = 11\sqrt{7} \)
In simple words: Pull out all perfect squares from under each radical sign to get like square roots, then add and subtract their outside numbers.

Exam Tip: Find the largest square factor (e.g. \( 36 \) in \( 252 \) and \( 49 \) in \( 294 \)) to reduce radicals in a single step.

Question 25. Give an example of two irrational numbers whose: (A) Sum is rational (B) product is rational (C) quotient is rational
Answer:
(A) Two irrational numbers whose sum is rational:
Let the numbers be \( 3 + \sqrt{2} \) and \( 3 - \sqrt{2} \).
Sum \( = (3 + \sqrt{2}) + (3 - \sqrt{2}) = 3 + 3 = 6 \), which is a rational number.

(B) Two irrational numbers whose product is rational:
Let the numbers be \( \sqrt{2} \) and \( \sqrt{8} \).
Product \( = \sqrt{2} \times \sqrt{8} = \sqrt{16} = 4 \), which is a rational number.

(C) Two irrational numbers whose quotient is rational:
Let the numbers be \( 2\sqrt{3} \) and \( \sqrt{3} \).
Quotient \( = \frac{2\sqrt{3}}{\sqrt{3}} = 2 \), which is a rational number.
In simple words: Conjugate pairs like 3 plus root 2 and 3 minus root 2 add to 6, roots like root 2 and root 8 multiply to 4, and 2 root 3 divided by root 3 cancels to 2.

Exam Tip: Using conjugate pairs of the form \( a + \sqrt{b} \) and \( a - \sqrt{b} \) is the most reliable way to show that operations on irrational numbers can yield rational results.