solve for x ifx^-(√3+1)x+√3=0
suvarnaalapathi:
please tell me how -√3came
Answers
Answered by
78
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
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
Answered by
27
Answer:
Step-by-step explanation:
Broo it's simple by finding roots by factorisation method.
Find it on the following image
Attachments:
Similar questions