y^2-4y-1=0 . solve quadratic equation
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y²-4y-1=0
using quadratic formula:
a = 1, b = -4, c = -1
∴ D = b² - 4ac = (-4)² - 4(1)(-1) = 20
20<0 in this case, the roots are real and distinct
let roots of given equation be α and β
α = -b+√d/2a
⇒ α = 4 + 2√2 / 2 =2(2+√2)/2= 2+√2
β = -b-√d/2a
⇒ β = 4 - 2√2 / 2 =2(2-√2)/2= 2-√2
hence roots of equation are 2+√2, 2-√2
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