form a quadratic equation one of whose roots is2+root 5&sum of the rootsis 4.?

Answer 1

If one root is 4, then (x-4) is a factor

I can't tell what your other root is, as the font is a little wacky from my end, but if you're saying the other root is (2+sqrt(5)), then (x-2+sqrt(5)) is your other factor

So take (x-4) * (x-2+sqrt(5)) = 0, then use FOIL (first, outside, inside, last) to multiply out your quadratic equation.

HTH! :-)

Answer 2

If one root is 2 + sqrt(5), the other must be 2 - sqrt(5). Obviously, they sum to 4.

So: (x - (2 + sqrt(5)))(x - (2 - sqrt(5))) = 0

x^2 - 4x - 1 = 0

Answer 3

(x - (2 + sqrt(5)))(x - (2 - sqrt(5)))

x^2 - (2 - sqrt(5))x - (2 + sqrt(5))x + ((2 + sqrt(5))(2 - sqrt(5))

x^2 - 2x + sqrt(5)x - 2x - sqrt(5)x + (4 - 5)

f(x) = x^2 - 4x - 1

Answer 4

the roots are 2+rt5 and 2-rt5
the equation
[x-(2+rt5)][x-(2-rt5)]=0