🧑‍🏫 Master Quadratic Equations, Logarithms & Systems — Teacher-Approved Guide

📐 MATHEMATICS SOLUTIONS | STEP-BY-STEP

1. Make x the subject: ax² + bx + c = 0

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Quadratic Formula Derivation

Starting from the standard form:

ax² + bx + c = 0

Step 1: Move c to the other side

ax² + bx = -c

Step 2: Divide both sides by a

x² + (b/a)x = -c/a

Step 3: Complete the square

x² + (b/a)x + (b/(2a))² = (b/(2a))² - c/a

Step 4: Simplify

(x + b/(2a))² = (b² - 4ac)/(4a²)

Step 5: Take square root

x + b/(2a) = ± √(b² - 4ac)/(2a)

Step 6: Solve for x

x = -b ± √(b² - 4ac) 2a

x = (-b ± √(b² - 4ac)) / 2a

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Final Answer

x = -b ± √(b² - 4ac)2a

2. Simplify: (x²√x)(√x∛x²) / (x³y³)^(1/2)

Step 1: Convert all radicals to powers

√x = x¹ᐟ²    ∛x² = x²ᐟ³

Step 2: Numerator:

x² · x¹ᐟ² = x52
x¹ᐟ² · x²ᐟ³ = x76
Numerator = x52 · x76 = x226 = x113

Step 3: Denominator:

(x³y³)¹ᐟ² = x³ᐟ² · y³ᐟ²

Step 4: Division:

x¹¹ᐟ³ x³ᐟ² · y³ᐟ² = x136 · y⁻³ᐟ²

x¹³ᐟ⁶ / y³ᐟ²

3. Solve: log(x+1) + log(x+1)² = 2 log(x+2)

Step 1: Use logarithm power rule: log A² = 2 log A

log(x+1) + 2 log(x+1) = 3 log(x+1)

Step 2: Rewrite equation

3 log(x+1) = 2 log(x+2)

Step 3: Divide by 3

log(x+1) = 23 log(x+2) = log(x+2)²ᐟ³

Step 4: Remove logs (assuming base 10)

x + 1 = (x + 2)²ᐟ³

Step 5: Cube both sides

(x + 1)³ = (x + 2)²

x³ + 3x² + 3x + 1 = x² + 4x + 4
x³ + 2x² - x - 3 = 0

Step 6: Domain: x > -1, x > -2 → x > -1

Step 7: Approximate solution by testing values

Test x = 1.5 → 3.375 + 4.5 - 1.5 - 3 = 3.375 (too high)
Test x = 1.2 → 1.728 + 2.88 - 1.2 - 3 = 0.408 (close)
Test x = 1.15 → 1.520875 + 2.645 - 1.15 - 3 = 0.015875 ≈ 0

x ≈ 1.15

4. Make r the subject: x/y = (1 + r²)/(1 - r²)

Step 1: Cross multiply

x(1 - r²) = y(1 + r²)

x - xr² = y + yr²

Step 2: Collect r terms

x - y = xr² + yr² = r²(x + y)

Step 3: Isolate r²

r² = x - yx + y

Step 4: Square root (assuming r positive)

r = √[(x - y)/(x + y)]

5. Solve the system of equations

2l₁ + 3l₂ - 4l₃ = 26 ... (1)

l₁ - 5l₂ - 3l₃ = -87 ... (2)

-7l₁ + 2l₂ + 6l₃ = 16 ... (3)

Step 1: Express l₁ from equation (2)

l₁ = 5l₂ + 3l₃ - 87

Step 2: Substitute into (1) and (3)

Equation (1):

2(5l₂ + 3l₃ - 87) + 3l₂ - 4l₃ = 26

10l₂ + 6l₃ - 174 + 3l₂ - 4l₃ = 26

13l₂ + 2l₃ = 200 ... (4)

Equation (3):

-7(5l₂ + 3l₃ - 87) + 2l₂ + 6l₃ = 16

-35l₂ - 21l₃ + 609 + 2l₂ + 6l₃ = 16

-33l₂ - 15l₃ = -593

33l₂ + 15l₃ = 593 ... (5)

Step 3: Solve (4) and (5)

From (4): l₂ = (200 - 2l₃)/13

Substitute into (5):

33·(200 - 2l₃)/13 + 15l₃ = 593

33(200 - 2l₃) + 195l₃ = 7709

6600 - 66l₃ + 195l₃ = 7709

129l₃ = 1109

l₃ ≈ 8.60

Step 4: Find l₂ and l₁

l₂ = (200 - 2×8.60)/13 ≈ 14.06

l₁ = 5(14.06) + 3(8.60) - 87 ≈ 9.09

l₁ ≈ 9.09,   l₂ ≈ 14.06,   l₃ ≈ 8.60

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📊 Final Answers Summary

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Boxed Answers

1. x = (-b ± √(b² - 4ac)) / 2a
2. x¹³ᐟ⁶ / y³ᐟ²
3. x ≈ 1.15
4. r = √[(x - y)/(x + y)]
5. l₁ ≈ 9.09, l₂ ≈ 14.06, l₃ ≈ 8.60

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📌 Summary: All five problems have been solved step by step. The quadratic formula was derived, the algebraic expression was simplified using exponent rules, the logarithmic equation was solved numerically, the formula was rearranged for r, and the linear system was solved using substitution.