Question

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

Answer 1

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

Answer 2

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

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

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