- Solve the simultaneous equations x + y = 2 (1)
- x2 − 10x + 17 (complete the square) (x2 − 10x + 5²) − 5² + 17
- Let's solve these two equations (system of equations) by substitution 2x + y = 3 (1)
- 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)
- f(x) = (2x + 3)(x + 4) (expand it) = 2x² − 5x − 7 ÷2 (divide it by 2)
-
- x² − (6√3)·x + 24 = 0 x² − (6√3)·x + (3√3)² − (3√3)² + 24 = 0
- x4 − (6√3)·x² + 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
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
⇒ (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).
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
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
= 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
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