## anonymous 4 years ago can someone explain this to me. verify that the numbers 1+sqrt 5 and 1- sqrt 5 both satisfy the equation x^2-2x-4=0

1. anonymous

let x = 1+sqrt(5) and see what happens. for example, (1+sqrt(5))^2 = (1+sqrt(5))(1+sqrt(5)) = 1+ 2sqrt(5) + 5 = 6+2sqrt(5) by distributing

2. ash2326

we have x^2-2x-4=0 let's find the roots of this one standard quadratic equation ax^2+bx+c=0 $x=(-b\pm\sqrt{b^2-4ac})/2a$ here a =1 b=-2 c=-4 $x=(2\pm\sqrt{4+16})/2$ or $x=1\pm\sqrt 5$ hence the two no. satisfy the ewuation

3. ash2326

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