d. A square piece of land has an area of x2 square metres. A smaller square of area 25 square metres is removed from one corner. Write an expression for the remaining area and factorise it. (3 marks)

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Asked on July 3, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-07-03T00:31:52.488Z

d. A square piece of land has an area of x^2 square metres. A smaller square of area 25 square metres is removed from one corner. Write an expression for the remaining area and factorise it. (3 marks) Step 1: Write the expression for the remaining area. The remaining area is the original area minus the removed area. Remaining Area = x^2 - 25 Step 2: Factorise the expression using the difference of two squares formula, A^2 - B^2 = (A - B)(A + B). Here, A=x and B=sqrt(25)=5. x^2 - 5^2 = (x - 5)(x + 5) The expression for the remaining area is x^2 - 25 and its factorised form is (x - 5)(x + 5). 3. a. Simplify the following (4 marks) i. (x)/(x+1) - (2x)/(x+2) Step 1: Find a common denominator, which is (x+1)(x+2). (x(x+2))/((x+1)(x+2)) - (2x(x+1))/((x+1)(x+2)) Step 2: Combine the fractions and expand the numerator. (x(x+2) - 2x(x+1))/((x+1)(x+2)) = (x^2 + 2x - (2x^2 + 2x))/((x+1)(x+2)) Step 3: Simplify the numerator. (x^2 + 2x - 2x^2 - 2x)/((x+1)(x+2)) = (-x^2)/((x+1)(x+2)) The simplified expression is (-x^2)/((x+1)(x+2)). ii. (4m^2 - 9)/((m-1)^2) ÷ (2m+3)/(m^2-m) Step 1: Factorise the numerators and denominators. 4m^2 - 9 = (2m - 3)(2m + 3) (difference of two squares) m^2 - m = m(m - 1) The expression becomes: ((2m - 3)(2m + 3))/((m-1)^2) ÷ (2m+3)/(m(m-1)) Step 2: Change division to multiplication by the reciprocal of the second fraction. ((2m - 3)(2m + 3))/((m-1)^2) × (m(m-1))/(2m+3) Step 3: Cancel common factors (2m+3) and (m-1). (2m - 3)(2m + 3)(m-1)(m-1) × m(m-1)2m+3 = (m(2m - 3))/(m-1) The simplified expression is (m(2m - 3))/(m-1). 3. b. For the fraction ((x-2)(x+3))/(x^2-4) i. Find the value(s) of x that make the fraction undefined. Explain your reasoning. (3 marks) Step 1: A fraction is undefined when its denominator is equal to zero. Set the denominator to zero: x^2 - 4 = 0 Step 2: Solve for x. x^2 = 4 x = ±sqrt(4) x = ± 2 The values of x that make the fraction undefined are x = 2 and x = -2. This is because these values make the denominator zero, which results in division by zero. ii. Find the value(s) of x that make the fraction zero. Explain your reasoning. (3 marks) Step 1: A fraction is zero when its numerator is equal to zero, provided the denominator is not also zero for that value of x. Set the numerator to zero: (x-2)(x+3) = 0 Step 2: Solve for x. x-2 = 0 x = 2 x+3 = 0 x = -3 Step 3: Check if these values make the denominator zero. For x=2, the denominator x^2-4 = 2^2-4 = 4-4=0. Since the denominator is zero, x=2 does not make the fraction zero; it makes it undefined. For x=-3, the denominator x^2-4 = (-3)^2-4 = 9-4=5. The denominator is not zero. The value of x that makes the fraction zero is x = -3. This is because when x=-3, the numerator is zero and the denominator is non-zero. 3. c. A company's profit is modelled by the fraction (2x)/(x-5) where x represents units sold. i. Explain what happens to the profit when x=5. Why is this significant? (2 marks) When x=5, the denominator x-5 becomes 5-5=0. This makes the profit expression (2(5))/(0) = (10)/(0), which is undefined. This is significant because it indicates a critical point where the mathematical model breaks down, possibly representing a scenario where the profit cannot be determined or a point of instability in the company's operations, such as a break-even point or a point of infinite loss/gain, depending on the context of the model. ii. Determine for which values of x the profit is positive. (2 marks) Step 1: For the profit to be positive, (2x)/(x-5) > 0. This occurs when both the numerator and denominator have the same sign. Case 1: Both are positive. 2x > 0 x > 0 x-5 > 0 x > 5 For both conditions to be true, x > 5. Case 2: Both are negative. 2x < 0 x < 0 x-5 < 0 x < 5 For both conditions to be true, x < 0. The profit is positive when x < 0 or x > 5. 4. a. Make the indicated variable the subject: (4 marks) i. Make y the subject of 3x + 2y = 12 Step 1: Subtract 3x from both sides. 2y = 12 - 3x Step 2: Divide both sides by 2. y = (12 - 3x)/(2) The subject is y = (12 - 3x)/(2). ii. Make r the subject of A = r^2 + 2 rh Step 1: Rearrange the equation into the standard quadratic form ar^2 + br + c = 0. r^2 + 2 hr - A = 0 Step 2: Use the quadratic formula r = -b ± sqrt(b^2 - 4ac)2a, where a=, b=2 h, and c=-A. r = -(2 h) ± sqrt((2 h)^2 - 4()(-A))2() Step 3: Simplify the expression. r = -2 h ± sqrt(4^2 h^2 + 4 A)2 r = -2 h ± sqrt(4( h^2 + A))2 r = -2 h ± 2sqrt(( h^2 + A))2 Step 4: Divide all terms by 2. r = -h ± sqrt(( h^2 + A)) The subject is r = -h ± sqrt(( h^2 + A)). 4. b. Solve the following equations: (6 marks) i. 5x - 3 = 2x + 9 Step 1: Subtract 2x from both sides. 5x - 2x - 3 = 9 3x - 3 = 9 Step 2: Add 3 to both sides. 3x = 9 + 3 3x = 12 Step 3: Divide by 3. x = (12)/(3) x = 4 The solution is x = 4. What's next?

Accepted Answer · 2026-07-03T00:31:52.488Z

d. A square piece of land has an area of x^2 square metres. A smaller square of area 25 square metres is removed from one corner. Write an expression for the remaining area and factorise it. (3 marks) Step 1: Write the expression for the remaining area. The remaining area is the original area minus the removed area. Remaining Area = x^2 - 25 Step 2: Factorise the expression using the difference of two squares formula, A^2 - B^2 = (A - B)(A + B). Here, A=x and B=sqrt(25)=5. x^2 - 5^2 = (x - 5)(x + 5) The expression for the remaining area is x^2 - 25 and its factorised form is (x - 5)(x + 5). 3. a. Simplify the following (4 marks) i. (x)/(x+1) - (2x)/(x+2) Step 1: Find a common denominator, which is (x+1)(x+2). (x(x+2))/((x+1)(x+2)) - (2x(x+1))/((x+1)(x+2)) Step 2: Combine the fractions and expand the numerator. (x(x+2) - 2x(x+1))/((x+1)(x+2)) = (x^2 + 2x - (2x^2 + 2x))/((x+1)(x+2)) Step 3: Simplify the numerator. (x^2 + 2x - 2x^2 - 2x)/((x+1)(x+2)) = (-x^2)/((x+1)(x+2)) The simplified expression is (-x^2)/((x+1)(x+2)). ii. (4m^2 - 9)/((m-1)^2) ÷ (2m+3)/(m^2-m) Step 1: Factorise the numerators and denominators. 4m^2 - 9 = (2m - 3)(2m + 3) (difference of two squares) m^2 - m = m(m - 1) The expression becomes: ((2m - 3)(2m + 3))/((m-1)^2) ÷ (2m+3)/(m(m-1)) Step 2: Change division to multiplication by the reciprocal of the second fraction. ((2m - 3)(2m + 3))/((m-1)^2) × (m(m-1))/(2m+3) Step 3: Cancel common factors (2m+3) and (m-1). (2m - 3)(2m + 3)(m-1)(m-1) × m(m-1)2m+3 = (m(2m - 3))/(m-1) The simplified expression is (m(2m - 3))/(m-1). 3. b. For the fraction ((x-2)(x+3))/(x^2-4) i. Find the value(s) of x that make the fraction undefined. Explain your reasoning. (3 marks) Step 1: A fraction is undefined when its denominator is equal to zero. Set the denominator to zero: x^2 - 4 = 0 Step 2: Solve for x. x^2 = 4 x = ±sqrt(4) x = ± 2 The values of x that make the fraction undefined are x = 2 and x = -2. This is because these values make the denominator zero, which results in division by zero. ii. Find the value(s) of x that make the fraction zero. Explain your reasoning. (3 marks) Step 1: A fraction is zero when its numerator is equal to zero, provided the denominator is not also zero for that value of x. Set the numerator to zero: (x-2)(x+3) = 0 Step 2: Solve for x. x-2 = 0 x = 2 x+3 = 0 x = -3 Step 3: Check if these values make the denominator zero. For x=2, the denominator x^2-4 = 2^2-4 = 4-4=0. Since the denominator is zero, x=2 does not make the fraction zero; it makes it undefined. For x=-3, the denominator x^2-4 = (-3)^2-4 = 9-4=5. The denominator is not zero. The value of x that makes the fraction zero is x = -3. This is because when x=-3, the numerator is zero and the denominator is non-zero. 3. c. A company's profit is modelled by the fraction (2x)/(x-5) where x represents units sold. i. Explain what happens to the profit when x=5. Why is this significant? (2 marks) When x=5, the denominator x-5 becomes 5-5=0. This makes the profit expression (2(5))/(0) = (10)/(0), which is undefined. This is significant because it indicates a critical point where the mathematical model breaks down, possibly representing a scenario where the profit cannot be determined or a point of instability in the company's operations, such as a break-even point or a point of infinite loss/gain, depending on the context of the model. ii. Determine for which values of x the profit is positive. (2 marks) Step 1: For the profit to be positive, (2x)/(x-5) > 0. This occurs when both the numerator and denominator have the same sign. Case 1: Both are positive. 2x > 0 x > 0 x-5 > 0 x > 5 For both conditions to be true, x > 5. Case 2: Both are negative. 2x < 0 x < 0 x-5 < 0 x < 5 For both conditions to be true, x < 0. The profit is positive when x < 0 or x > 5. 4. a. Make the indicated variable the subject: (4 marks) i. Make y the subject of 3x + 2y = 12 Step 1: Subtract 3x from both sides. 2y = 12 - 3x Step 2: Divide both sides by 2. y = (12 - 3x)/(2) The subject is y = (12 - 3x)/(2). ii. Make r the subject of A = r^2 + 2 rh Step 1: Rearrange the equation into the standard quadratic form ar^2 + br + c = 0. r^2 + 2 hr - A = 0 Step 2: Use the quadratic formula r = -b ± sqrt(b^2 - 4ac)2a, where a=, b=2 h, and c=-A. r = -(2 h) ± sqrt((2 h)^2 - 4()(-A))2() Step 3: Simplify the expression. r = -2 h ± sqrt(4^2 h^2 + 4 A)2 r = -2 h ± sqrt(4( h^2 + A))2 r = -2 h ± 2sqrt(( h^2 + A))2 Step 4: Divide all terms by 2. r = -h ± sqrt(( h^2 + A)) The subject is r = -h ± sqrt(( h^2 + A)). 4. b. Solve the following equations: (6 marks) i. 5x - 3 = 2x + 9 Step 1: Subtract 2x from both sides. 5x - 2x - 3 = 9 3x - 3 = 9 Step 2: Add 3 to both sides. 3x = 9 + 3 3x = 12 Step 3: Divide by 3. x = (12)/(3) x = 4 The solution is x = 4. What's next?

d. A square piece of land has an area of x2 square metres. A smaller square of area 25 square metres is removed from one corner. Write an expression for the remaining area and factorise it. (3 marks)

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d. A square piece of land has an area of x2 square metres. A smaller square of area 25 square metres is removed from one corner. Write an expression for the remaining area and factorise it. (3 marks)

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