Question

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

Answer 1

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

Answer 2

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

Answer 3

*equation