Write the quadratic equation whose coefficient with x^2 to 1 and roots are √3-1 divided by 2 and √3+1 divided by 2

Now that we have both of these zero terms, we can multiply them to get a standard form.

f (x) = (x - 6) (x + 4)

And while this will give us the zeros we need, it will no give us the lead coefficient. So we must multiply by the desired lead coefficient.

f (x) = 5 (x - 6) (x + 4)

f (x) = 5 (x^2 - 6x + 4x - 24)

f (x) = 5 (x^2 - 2x - 24)

f (x) = 5x^2 - 10x - 120

1x^2-sqrt (3) b+1/2=0 or

2x^2 - 2sqrt (3) b+1=0

Step-by-step explanation:

ax^2 + bx+c=0

a=1, b, c=?

x1 = (sqrt (3) - 1) / 2

x2 = (sqrt (3) + 1) / 2

x1+x2=-b/a

x1 * x2 = c/a

x1 + x2=

(sqrt (3) - 1) / 2 + (sqrt (3) + 1) / 2=

2sqrt (3) / 2=

sqrt (3) = -b/1

b=-sqrt (3)

x1*x2 =

((sqrt (3) - 1) / 2) * ((sqrt (3) + 1) / 2) =

((sqrt (3)) ^2 - 1^2) / 4=

(3-1) / 4=

2/4=

1/2=c/1

c=1/2

ax^2 + bx+c=0

1x^2-sqrt (3) b+1/2=0 / *2

2x^2 - 2sqrt (3) b+1=0