Example 1 :
Solve by completing the square method
x2 – (√3 + 1)x + √3 = 0
Solution :
x2 – (√3 + 1)x + √3 = 0
x2 – (√3 + 1)x = -√3
x2 – (√3 + 1)x + [(√3 + 1)/2]2 = -√3 + [(√3 + 1)/2]2
[x - (√3 + 1)/2]2 = -√3 + [(√3 + 1)2 / 4]
[x - (√3 + 1)/2]2 = (-4√3 + √32 + 2√3 + 1)/4
[x - (√3 + 1)/2]2 = (√32 - 2√3 + 1)/4
[x - (√3 + 1)/2]2 = [ (√3 - 1)/2 ]2
[x - (√3 + 1)/2]2 = [ (√3 - 1)/2 ]2
[x - (√3 + 1)/2] = ± (√3 - 1)/2
x - (√3+1)/2 = (√3-1)/2 x = (√3-1)/2 + (√3+1)/2 x = 2√3/2 x = √3 |
x - (√3+1)/2 = -(√3-1)/2 x = -(√3-1)/2 + (√3+1)/2 x = 2/2 x = 1 |
Example 2 :
Solve by completing the square method
(5x + 7)/(x – 1) = 3x + 2
Solution :
(5x + 7) = (3x + 2)(x – 1)
5x + 7 = 3x² – 3x + 2x – 2
3x² – 3x + 2x – 2 – 5x – 7 = 0
3x² – 6x – 9 = 0
Divide the entire equation by 3, we get
x² – 2x – 3 = 0
x² – 3x + x – 3 = 0
x (x – 3) + 1 (x – 3) = 0
(x – 3)(x + 1) = 0
x - 3 = 0 x = 3 |
x + 1 = 0 x = -1 |
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