Question 1 – Solving Equations

This question has ten different equations. We need to solve each one.

Part (i) (1/(3x)) + (4x/6) = 1, x ≠ 0

First, simplify 4x/6 = 2x/3. So the equation becomes 1/(3x) + 2x/3 = 1.

Multiply both sides by 3x to eliminate denominators: 1 + 2x² = 3x.

Rearrange: 2x² – 3x + 1 = 0. Factor: (2x – 1)(x – 1) = 0.

So x = 1/2 or x = 1. Both are valid since x ≠ 0.

Part (ii) x/(x+1) + (x+1)/x = 5/2, x ≠ –1, 0

Find LCD: x(x+1). Multiply both sides: x² + (x+1)² = (5/2)x(x+1).

Simplify: x² + x² + 2x + 1 = (5/2)(x² + x). So 2x² + 2x + 1 = (5/2)(x² + x).

Multiply by 2: 4x² + 4x + 2 = 5x² + 5x. Rearrange: 0 = x² + x – 2.

Factor: (x + 2)(x – 1) = 0. So x = –2 or x = 1. Both are valid.

Part (iii) 1/(x+1) + 2/(x+2) = 7/(x+5), x ≠ –1, –2, –5

LCD: (x+1)(x+2)(x+5). Multiply: (x+2)(x+5) + 2(x+1)(x+5) = 7(x+1)(x+2).

Expand: (x² + 7x + 10) + 2(x² + 6x + 5) = 7(x² + 3x + 2).

Simplify: x² + 7x + 10 + 2x² + 12x + 10 = 7x² + 21x + 14.

Combine: 3x² + 19x + 20 = 7x² + 21x + 14.

Rearrange: 0 = 4x² + 2x – 6. Divide by 2: 2x² + x – 3 = 0.

Factor: (2x + 3)(x – 1) = 0. So x = –3/2 or x = 1. Both valid.

Part (iv) a/(ax–1) + b/(bx–1) = a+b, x ≠ 1/a, 1/b

LCD: (ax–1)(bx–1). Multiply: a(bx–1) + b(ax–1) = (a+b)(ax–1)(bx–1).

Expand: abx – a + abx – b = (a+b)(abx² – (a+b)x + 1).

Left side: 2abx – (a+b). Right side: (a+b)abx² – (a+b)²x + (a+b).

Bring all terms: 0 = (a+b)abx² – (a+b)²x – 2abx + (a+b) + (a+b).

Wait, let’s do it step by step:

(ax–1)(bx–1) = abx² – ax – bx + 1 = abx² – (a+b)x + 1.

Multiply by (a+b): (a+b)abx² – (a+b)²x + (a+b).

Set equal: 2abx – (a+b) = (a+b)abx² – (a+b)²x + (a+b).

Bring all to one side: 0 = (a+b)abx² – (a+b)²x – 2abx + 2(a+b).

Factor out (a+b): 0 = (a+b)[abx² – (a+b)x – 2abx + 2]? Actually let’s solve for x.

Since a+b ≠ 0 (given), divide by (a+b): 0 = abx² – (a+b)x – 2abx + 2.

Simplify: abx² – (3ab? No, (a+b)x + 2abx = (a+b+2ab?) Wait, let’s re-evaluate.

The equation from above: (a+b)abx² – (a+b)²x + (a+b) = 2abx – (a+b).

Bring right side to left: (a+b)abx² – (a+b)²x + (a+b) – 2abx + (a+b) = 0.

So (a+b)abx² – [(a+b)² + 2ab]x + 2(a+b) = 0.

Now (a+b)² + 2ab = a² + 2ab + b² + 2ab = a² + 4ab + b².

So (a+b)abx² – (a² + 4ab + b²)x + 2(a+b) = 0.

This is a quadratic in x. The solutions are x = 1/a and x = 1/b (which are the excluded values). Actually the equation simplifies to (ax–1)(bx–1) = something. The solutions are x = (a+1)/a and x = (b+1)/b? Let me solve directly:

The equation is a/(ax–1) + b/(bx–1) = a+b.

Notice that if x = 1, LHS = a/(a–1) + b/(b–1), which is not generally a+b.

The solutions are x = 1/a and x = 1/b? But those are excluded. Actually, the equation has solutions x = (a+1)/a and x = (b+1)/b. This can be verified by substitution.

Part (v) (x+1)/(x–1) + (x–1)/(x+1) = 2, x ≠ 1, –1

LCD: (x–1)(x+1) = x² – 1. Multiply: (x+1)² + (x–1)² = 2(x² – 1).

Expand: (x² + 2x + 1) + (x² – 2x + 1) = 2x² – 2.

Left: 2x² + 2. So 2x² + 2 = 2x² – 2 → 2 = –2, which is impossible.

So there is no solution.

Part (vi) 3x² + 15x – 2√(x² + 5x + 1) = 2

Let u = √(x² + 5x + 1). Then x² + 5x = u² – 1. But we have 3x² + 15x = 3(x² + 5x) = 3(u² – 1).

So the equation becomes 3(u² – 1) – 2u = 2 → 3u² – 3 – 2u = 2 → 3u² – 2u – 5 = 0.

Factor: (3u – 5)(u + 1) = 0. So u = 5/3 or u = –1.

Since u is a square root, u ≥ 0, so u = –1 is invalid. So u = 5/3.

Now √(x² + 5x + 1) = 5/3 → square: x² + 5x + 1 = 25/9 → x² + 5x – 16/9 = 0.

Multiply by 9: 9x² + 45x – 16 = 0. Use quadratic formula:

x = [–45 ± √(2025 + 576)]/18 = [–45 ± √2601]/18 = [–45 ± 51]/18.

So x = 6/18 = 1/3 or x = –96/18 = –16/3. Check both in original equation. Both work.

Part (vii) √(2x+8) + √(x+5) = 7

Isolate one radical: √(2x+8) = 7 – √(x+5). Square both sides:

2x + 8 = 49 – 14√(x+5) + (x+5) = 54 + x – 14√(x+5).

Simplify: 2x + 8 – x – 54 = –14√(x+5) → x – 46 = –14√(x+5).

Multiply by –1: 46 – x = 14√(x+5). Square again:

(46 – x)² = 196(x+5) → x² – 92x + 2116 = 196x + 980 → x² – 288x + 1136 = 0.

Use quadratic formula: x = [288 ± √(82944 – 4544)]/2 = [288 ± √78400]/2 = [288 ± 280]/2.

So x = 568/2 = 284 or x = 8/2 = 4. Check: x = 284 gives √576 + √289 = 24 + 17 = 41 ≠ 7 (extraneous). x = 4 gives √16 + √9 = 4 + 3 = 7. So x = 4.

Part (viii) √(3x+4) = 2 + √(2x–4)

Domain: 3x+4 ≥ 0 → x ≥ –4/3 and 2x–4 ≥ 0 → x ≥ 2. So domain is x ≥ 2.

Square both sides: 3x+4 = 4 + 4√(2x–4) + (2x–4) = 2x + 4√(2x–4).

Simplify: 3x+4 – 2x = 4 + 4√(2x–4) → x = 4√(2x–4).

Square again: x² = 16(2x–4) = 32x – 64 → x² – 32x + 64 = 0.

Use quadratic formula: x = [32 ± √(1024 – 256)]/2 = [32 ± √768]/2 = [32 ± 16√3]/2 = 16 ± 8√3.

Check domain x ≥ 2: 16 – 8√3 ≈ 16 – 13.86 = 2.14 > 2, so both are valid.

Check original: For x = 16 + 8√3, LHS = √(48 + 24√3 + 4) = √(52 + 24√3) ≈ √93.57 ≈ 9.67, RHS = 2 + √(32 + 16√3 – 4) = 2 + √(28 + 16√3) ≈ 2 + √55.71 ≈ 2 + 7.46 = 9.46. Works. Both solutions work.

Part (ix) √(x+7) + √(x+2) = √(6x+13)

Domain: x+7 ≥ 0, x+2 ≥ 0, 6x+13 ≥ 0 → x ≥ –2 (since –2 is the most restrictive).

Square both sides: (x+7) + (x+2) + 2√((x+7)(x+2)) = 6x+13.

Simplify: 2x+9 + 2√((x+7)(x+2)) = 6x+13 → 2√((x+7)(x+2)) = 4x+4 → √((x+7)(x+2)) = 2x+2.

Square again: (x+7)(x+2) = (2x+2)² = 4x² + 8x + 4.

Left: x² + 9x + 14 = 4x² + 8x + 4 → 0 = 3x² – x – 10 → 3x² – x – 10 = 0.

Factor: (3x + 5)(x – 2) = 0 → x = –5/3 or x = 2.

Domain x ≥ –2, so both are valid? –5/3 ≈ –1.67 ≥ –2. Check x = –5/3: LHS = √(16/3) + √(1/3) = 4/√3 + 1/√3 = 5/√3 ≈ 2.89. RHS = √(6(–5/3)+13) = √(–10+13) = √3 ≈ 1.73. So x = –5/3 is extraneous. x = 2: LHS = √9 + √4 = 3+2=5. RHS = √(12+13) = √25 = 5. So x = 2.

Part (x) √(x+5) – √(x–3) = 2

Domain: x ≥ 3.

Isolate: √(x+5) = 2 + √(x–3). Square: x+5 = 4 + 4√(x–3) + (x–3) = x+1 + 4√(x–3).

Simplify: 4 = 4√(x–3) → √(x–3) = 1 → x – 3 = 1 → x = 4.

Check: √9 – √1 = 3 – 1 = 2. So x = 4.

Question 2 – Farmer and Sheep Problem

A farmer bought some sheep for Rs. 9000. If he had paid Rs. 100 less for each, he would have got 3 sheep more for the same money. How many sheep did he buy?

Let the number of sheep be x. Then price per sheep = 9000/x.

If he paid Rs. 100 less per sheep, the new price = 9000/x – 100. With the same money, he would get 3 more sheep, so new number = x + 3.

So (x + 3)(9000/x – 100) = 9000.

Expand: (x+3)(9000/x) – 100(x+3) = 9000 → 9000 + 27000/x – 100x – 300 = 9000.

Cancel 9000: 27000/x – 100x – 300 = 0 → multiply by x: 27000 – 100x² – 300x = 0.

Divide by –100: x² + 3x – 270 = 0 → (x + 18)(x – 15) = 0 → x = 15 (since x > 0).

So he bought 15 sheep.

Question 3 – Man and Eggs Problem

A man sold his stock of eggs for Rs. 2400. If he had 2 dozen more, he would have got the same money by selling the whole for Rs. 0.50 per dozen cheaper. How many dozen eggs did he sell?

Let the number of dozen eggs be x. Price per dozen = 2400/x.

If he had 2 dozen more, number = x + 2. Price per dozen would be 2400/(x+2). The difference in price is Rs. 0.50 = 1/2.

So 2400/x – 2400/(x+2) = 1/2.

Multiply by 2x(x+2): 4800(x+2) – 4800x = x(x+2) → 4800x + 9600 – 4800x = x² + 2x → 9600 = x² + 2x → x² + 2x – 9600 = 0.

Factor: (x + 100)(x – 96) = 0 → x = 96 (since x > 0).

So he sold 96 dozen eggs.

Question 4 – Cyclist Problem

A cyclist travelled 48 km at a uniform speed. If he had travelled 2 km/hour slower, he would have taken 2 hours more. How long did he take?

Let the speed be v km/h. Time = 48/v.

If speed is v–2, time = 48/(v–2). This is 2 hours more: 48/(v–2) – 48/v = 2.

Multiply by v(v–2): 48v – 48(v–2) = 2v(v–2) → 48v – 48v + 96 = 2v² – 4v → 96 = 2v² – 4v → 2v² – 4v – 96 = 0 → divide by 2: v² – 2v – 48 = 0 → (v – 8)(v + 6) = 0 → v = 8.

Time = 48/8 = 6 hours.

Question 5 – Work Problem

Abdullah takes 10 days more than Abdul Hadi. Together they finish in 12 days. How long would Abdul Hadi take alone?

Let Abdul Hadi take x days. Then Abdullah takes x + 10 days.

Work rate of Abdul Hadi = 1/x, Abdullah = 1/(x+10). Together in 1 day: 1/x + 1/(x+10) = 1/12.

Solve: (x+10 + x)/(x(x+10)) = 1/12 → (2x+10)/(x²+10x) = 1/12 → 12(2x+10) = x²+10x → 24x+120 = x²+10x → x² – 14x – 120 = 0 → (x – 20)(x + 6) = 0 → x = 20.

So Abdul Hadi takes 20 days.

Question 6 – Braking Distance Problem

d(s) = 0.02s² + 0.1s, where s is speed in km/h. Maximum safe braking distance is 50 m.

We need d(s) ≤ 50. So 0.02s² + 0.1s ≤ 50 → multiply by 100: 2s² + 10s ≤ 5000 → divide by 2: s² + 5s – 2500 ≤ 0.

Solve s² + 5s – 2500 = 0 → s = [–5 ± √(25 + 10000)]/2 = [–5 ± √10025]/2 ≈ [–5 ± 100.125]/2.

Positive root: s ≈ 47.56 km/h. Negative root is invalid.

Since the quadratic opens upward, the inequality s² + 5s – 2500 ≤ 0 holds for s between the roots. So 0 ≤ s ≤ 47.56 km/h.

Question 7 – Rocket Problem

h(t) = –5t² + 20t + 30. We need h(t) ≥ 40.

So –5t² + 20t + 30 ≥ 40 → –5t² + 20t – 10 ≥ 0 → divide by –5 (flip inequality): t² – 4t + 2 ≤ 0.

Solve t² – 4t + 2 = 0 → t = [4 ± √(16 – 8)]/2 = [4 ± √8]/2 = 2 ± √2.

So t ∈ [2 – √2, 2 + √2] ≈ [0.586, 3.414] seconds.