Solving equations - miscellaneous exercices

Possible Solutions

1. Solve the simultaneous equations
    x + y = 2 (1)
    x² + 2y² = 11 (2)

It's a system of equations, consisting of a linear and a quadratic equation. We will solve it by Substitution, the only method that suits here.
Important: It really doesn't matter what equation or what variable you pick, the answer will the same. "Kinda vice versa" ...cool, ain't it?

- first solve (1) for y:
⇒ y = 2 − x (you can call it (3), just to put everything in order)

- then plug (3) into (2)
⇒ x² + 2(2 − x)² = 11 (you can notice only one unknown variable left, makes it easier!)

- expand:
⇒ x² + 2(4 −4x + x²) = 11 (take your time and you'll come to the following result)
⇔ 3x² − 8x − 3 = 0  |  ÷3 (divide the whole equation by 3, on both side please!)
⇒ x² − ⁸/₃x − 1 = 0 (quadratic form)

- now you can solve it for: (by completing the square or what ever method you know!)
⇒ x² − ⁸/₃x + (⁴/₃)² − 1 − (⁴/₃)²
⇒ (x − ⁴/₃)² = 1 + ¹⁶/₉ = ²⁵/₉ (square root it)
⇒ x − ⁴/₃ = ± ⁵/₃
⇔ x₁ = − ⁵/₃ + ⁴/₃ = − ¹/₃,     x₁ = − ¹/₃
     or
     x₂ = ⁵/₃ + ⁴/₃ = 3,     x₂ = 3

- then you can replace both value of x in the equation (3) to find the values of y:
⇔ y₁ = 2 − (−¹/₃) = 2 + ¹/₃ = ⁷/₃,     y₁ = ⁷/₃
    y₂ = 2 − 3 = − 1,     y₂ = − 1

⇒ Then the solution is (x, y) = (−¹/₃ , ⁷/₃) or (3, −1).

2. x² − 10x + 17 (complete the square)
(x² − 10x + 5²) − 5² + 17
⇒ (x − 5)² − 8, with a = − 5 and b = − 8.
The least possible value that f(x) can take is − 8 (f(x) = y = −8) and its corresponding x value is − 5 (x = − 5).


3. Let's solve these two equations (system of equations) by substitution
    2x + y = 3 (1)
    2x² - xy = 10 (2)

- first solve (1) for y (kinda isolating y)
    ⇒ y = 3 − 2x (3)

- plug (3) into (2)
    2x² −x(3 − 2x) = 10 (expand it)
    2x² − 3x + 2x² = 10
    4x² − 3x − 10 = 0 (factorize 4)
    4(x² − ³/₄x − ¹⁰/₄) = 0 (complete the square)
    (x − ³/₄x + ³/₈)² − (³/₈)² − ¹⁰/₄
       (x − ³/₈)² = ^13²/_8² (square root it)
       x − ³/₈ = ± ¹³/₈

⇒
       x₁ = − ¹³/₈ + ³/₈ = − ⁵/₄,     x₁ = − ⁵/₄
         or
       x₂ = ¹³/₈ + ³/₈ = 2,    x₂ = 2

Substitute the values of x₁ and x₂ in (3) (kinda pluggin back into...) to find the two values of y:
Do you remember y = 3 − 2x (3)? ...great just askink! :-)

what gives u:
                 - for x = − ⁵/₄,          y = 3 − 2(− ⁵/₄) = ¹¹/₂
                 - for x = 2,               y = 3 − 2(2) = − 1

⇒ Then the solution is (x, y) = (−⁵/₄ , ¹¹/₂) or (2, −1).

4. 2x² − kx + 8 = 0 (divide it by 2 to bring it to a regular quadratic equation, without the factor 2 at x²)
⇒ x² − ^k/₂·x + 4 = 0 (complete the square)
          x² − ^k/₂·x + (^k/₄·x)² − (^k/₄·x)² + 4 = 0
          x² − ^k/₂·x + (^k/₄·x)² = (^k/₄·x)² - 4
Now compare this two equations:
     x² − ^k/₂x + 4 = 0
         and
     x² − ^k/₂·x + (^k/₄·x)² = (^k/₄·x)² - 4

as you can see by identifying:
x² = x² is a true statement;
it goes the same for − ^k/₂·x = − ^k/₂·x
(^k/₄)² = 4
and (^k/₄)² - 4 = 0 impossible!

Therefore you can work out: (^k/₄)² = 4  |   √ (square root it)
⇒ ^k/₄ = ± 2 and k = ± 8

★
You can also use the Discriminant (Δ):
The equation again: 2x² − kx + 8 = 0
by identification:
     Δ = b² − 4·a·c
with a = 2; b = −k and c = 8
Δ = k² − 4 · 2 · 8 = 0
⇒ k² − 64 = 0
⇒ k² = 64   |   √ (square root it)
     what gives you: k = ± 8


5. f(x) = (2x + 3)(x + 4)     (expand it)
      = 2x² − 5x − 7    ÷2 (divide it by 2)
      = x² − ⁵/₂·x − ⁷/₂       (complete the square)
      = x² − ⁵/₂·x + (⁵/₄)² − (⁵/₄)² − ⁷/₂
by working it out, you'll get:
(x − ⁵/₄)² = ⁸¹/₁₆   |   √ (square root it)
⇒ x = ⁵/₄ ± ⁹/₄ ⇒ x = ⁵/₄ − ⁹/₄ = − 1 or x = ⁵/₄ + ⁹/₄ = ⁷/₂
           x = − 1
                or
           x = ⁷/₂


6. 1. x² − (6√3)·x + 24 = 0
   x² − (6√3)·x + (3√3)² − (3√3)² + 24 = 0
   work it out
   to get: (x − 3√3)² = 3   |   √ (square root it)
   ⇒ x − 3√3 = ± √3
   
   Therefore the answer in terms of surds is:
              x = 2√3
              or
              x = 4√3
   
   
   2. x⁴ − (6√3)·x² + 24 = 0

Coming soon...