Number System — Practice Quiz

2

Step 1: We need to simplify (√5 + √3)(√5 - √3). Step 2: This is of the form (a + b)(a - b) = a² - b², where a = √5 and b = √3. Step 3: Apply the identity: (√5)² - (√3)² = 5 - 3 = 2. Step 4: The result is 2, which is a rational number! This shows that the product of two irrational numbers can sometimes be rational. Step 5: Common mistake — students often try to expand term by term and make errors. Always look for the (a+b)(a-b) = a²-b² pattern first. The answer is 2.

√7 + √6

Step 1: To rationalise 1/(√7 - √6), multiply numerator and denominator by the conjugate (√7 + √6). Step 2: Numerator becomes: 1 × (√7 + √6) = √7 + √6. Step 3: Denominator becomes: (√7 - √6)(√7 + √6) = (√7)² - (√6)² = 7 - 6 = 1. Step 4: So the result = (√7 + √6)/1 = √7 + √6. Step 5: The key insight is that (√7 - √6)(√7 + √6) = 7 - 6 = 1, making the denominator 1. This is a special case where the denominator becomes exactly 1 after rationalisation. Common mistake: students choose (√7 + √6)/13, confusing this with cases where the denominator gives a value like 49 - 36 = 13.

4

Step 1: Given x = 2 + √3, we need to find x + 1/x. Step 2: First find 1/x = 1/(2 + √3). Rationalise by multiplying by (2 - √3)/(2 - √3). Step 3: 1/x = (2 - √3)/((2 + √3)(2 - √3)) = (2 - √3)/(4 - 3) = (2 - √3)/1 = 2 - √3. Step 4: Now compute x + 1/x = (2 + √3) + (2 - √3) = 2 + √3 + 2 - √3 = 4. Step 5: The irrational parts cancel out perfectly, giving a rational answer of 4. This demonstrates an important property: when x = a + √b, then 1/x = a - √b (after rationalisation when a² - b = 1), and x + 1/x = 2a. Here 2a = 2×2 = 4. ✓

2

Step 1: We need to simplify 2^(2/3) × 2^(1/3). Step 2: Use the law of exponents: a^m × a^n = a^(m+n). Here the base is the same (base = 2). Step 3: Add the exponents: 2/3 + 1/3 = 3/3 = 1. Step 4: So 2^(2/3) × 2^(1/3) = 2^1 = 2. Step 5: Common mistake — students multiply the exponents (2/3 × 1/3 = 2/9) instead of adding them. The rule a^m × a^n = a^(m+n) applies when multiplying same bases. Multiplying exponents (a^m)^n = a^(mn) applies when raising a power to another power. The answer is 2.