Solve : x2-(root3+1)x+root3=0 by the method of completing the square.

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• -18
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• 9
x² - (√3 + 1)x + √3 = 0

x² - 2(√3 + 1)(x)/2 + √3 = 0

x² - 2(√3 + 1)x/2 + (√3 + 1)²/2² - (√3 + 1)²/2² + √3 = 0

[x - (√3 + 1)/2]² - (3 + 1 + 2√3)/4 + √3 = 0

[x - (√3 + 1)/2]² = (3 + 1 + 2√3 - 4√3)/4

[x - (√3 + 1)/2]² = (√3 - 1)²/2²

[x - (√3 + 1)/2] = $$\frac{+}{}$$(√3 - 1)/2

taking (+ve)

x = (√3 - 1)/2 + (√3 + 1)/2

x = (√3 - 1 + √3 + 1)/2

x = 2√3/2 = √3

taking (-ve)

x = -(√3 - 1)/2 + (√3 + 1)/2

x = (-√3 + 1 + √3 + 1)/2

x = 2/2 = 1
• 21
x² - (√3 + 1)x + √3 = 0 x² - 2(√3 + 1)(x)/2 + √3 = 0 x² - 2(√3 + 1)x/2 + (√3 + 1)²/2² - (√3 + 1)²/2² + √3 = 0 [x - (√3 + 1)/2]² - (3 + 1 + 2√3)/4 + √3 = 0 [x - (√3 + 1)/2]² = (3 + 1 + 2√3 - 4√3)/4 [x - (√3 + 1)/2]² = (√3 - 1)²/2² [x - (√3 + 1)/2] = ±(√3 - 1)/2 taking (+ve) x = (√3 - 1)/2 + (√3 + 1)/2 x = (√3 - 1 + √3 + 1)/2 x = 2√3/2 = √3 taking (-ve) x = -(√3 - 1)/2 + (√3 + 1)/2 x = (-√3 + 1 + √3 + 1)/2 x = 2/2 = 1
• -2

x² - (√3 + 1)x + √3 = 0

x² - 2(√3 + 1)(x)/2 + √3 = 0

x² - 2(√3 + 1)x/2 + (√3 + 1)²/2² - (√3 + 1)²/2² + √3 = 0

[x - (√3 + 1)/2]² - (3 + 1 + 2√3)/4 + √3 = 0

[x - (√3 + 1)/2]² = (3 + 1 + 2√3 - 4√3)/4

[x - (√3 + 1)/2]² = (√3 - 1)²/2²

[x - (√3 + 1)/2] = $$\frac{+}{}$$(√3 - 1)/2

taking (+ve)

x = (√3 - 1)/2 + (√3 + 1)/2

x = (√3 - 1 + √3 + 1)/2

x = 2√3/2 = √3

taking (-ve)

x = -(√3 - 1)/2 + (√3 + 1)/2

x = (-√3 + 1 + √3 + 1)/2

x = 2/2 = 1
• -1
x? - (?3 + 1)x + ?3 = 0

x? - 2(?3 + 1)(x)/2 + ?3 = 0

x? - 2(?3 + 1)x/2 + (?3 + 1)?/2? - (?3 + 1)?/2? + ?3 = 0

[x - (?3 + 1)/2]? - (3 + 1 + 2?3)/4 + ?3 = 0

[x - (?3 + 1)/2]? = (3 + 1 + 2?3 - 4?3)/4

[x - (?3 + 1)/2]? = (?3 - 1)?/2?

[x - (?3 + 1)/2] = $$frac{+}{}$$(?3 - 1)/2

taking (+ve)

x = (?3 - 1)/2 + (?3 + 1)/2

x = (?3 - 1 + ?3 + 1)/2

x = 2?3/2 = ?3

taking (-ve)

x = -(?3 - 1)/2 + (?3 + 1)/2

x = (-?3 + 1 + ?3 + 1)/2

x = 2/2 = 1

• 2
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