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secondary 3 | E Maths
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Date Posted: 3 years ago
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If the roots are 2 - √3 and 2 + √3 , then x = 2 - √3 and x = 2 + √3 are solutions to the quadratic equation.
So x - (2 - √3) = 0 and x - (2 + √3) = 0
Multiplying both still gives 0.
(x - (2 - √3))(x - (2 + √3)) = 0
x² - (2 + √3)x - (2 - √3)x + (2 - √3)(2 + √3) = 0
x² - 2x - √3 x - 2x + √3 x + (2² - (√3)²) = 0
x² - 4x + (4 - 3) = 0
x² - 4x + 1 = 0 (the required equation)
Alternatively,
Sum of roots = 2 - √3 + 2 + √3 = 4
Product of roots = (2 - √3)(2 + √3)
= 2² - (√3)² = 4 - 3 = 1
Quadratic equation is : x² - (sum)x + product = 0
x² - 4x + 1 = 0
So x - (2 - √3) = 0 and x - (2 + √3) = 0
Multiplying both still gives 0.
(x - (2 - √3))(x - (2 + √3)) = 0
x² - (2 + √3)x - (2 - √3)x + (2 - √3)(2 + √3) = 0
x² - 2x - √3 x - 2x + √3 x + (2² - (√3)²) = 0
x² - 4x + (4 - 3) = 0
x² - 4x + 1 = 0 (the required equation)
Alternatively,
Sum of roots = 2 - √3 + 2 + √3 = 4
Product of roots = (2 - √3)(2 + √3)
= 2² - (√3)² = 4 - 3 = 1
Quadratic equation is : x² - (sum)x + product = 0
x² - 4x + 1 = 0
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