Question

Determine the equation of the parabola with roots 2+sqrt(3) and 2-sqrt(3), and passing through the point (2,5).

Answer 1

y=(x-r1)(x-r2) +b
= (x-2-sqrt3)(x-2+sqrt3)+b
= (x^2 -1) + b

for the point 2,5
5=(4-1)+b
solve for b, and you have it.

Answer 2

I respectfully must disagree with Bob

Since we know the roots, we can say the equation is

y = a(x-2-sqrt(3))(x-2+sqrt(3))
y = a(x^2 - 4x + 1)
plug in (2,5)
5 = a(4-8+1)
a= -5/3

equation: y = (-5/3)(x^2 - 4x + 1)
or y = (-5/3)(x-2)^2 + 5 after completing the square.

check:
#1: solving (-5/3)(x-2)^2 + 5 yiels 2 ± sqrt(3) , and
#2: (2,5) satisfies the equation.

Answer 3

Reiny is correct. I did it in my head, and missed the second term. Ugh.

Answer 4

Given: Ints 2 + sqrt(3) , 2 - sqrt(3).

x = 2 + sqrt(3),
x - 2 = sqrt(3),
Square both sides:
x^2 - 4x + 4 = 3,
Set Eq = 0:
x^2 - 4x + 1 = 0,

EQ: Y = X^2 - 4X + 1.

h = Xv = -a/b =4/2 = 2,
K = Yv = 2^2 - 4*2 + 1 - -3
V(2 , -3).


CHECK: y = (2 - sqrt(3)^2 -4(2 - sqrt(3)) + 1,
y = 4 - 4*sqrt(3) + 3 -8 + 4*sqrt(3)
+ 1,
Combine like-terms:
4 + 3 - 8 + 1 = 0. Ints satisfies EQ.

The X-Ints nd vertex satisfy the Eq, but the given point do not. Please
check the point for error.

Answer 5

Thank you!

Answer 6

Henry, the point (2,5) does not satisfy your equation.

The point (2,5) is the vertex, not (2,-3)

Answer 7

y=a(x - 2 - sqrt(3))(x - 2 + sqrt(3))
FOIL = First Outsisde Inside Last
y=a(x^2 - 2x + 1x - 2x + 4 - 2sqrt(3) - 1x + 2sqrt(3) - sqrt(9))
y=a(x^2 - 2x + 1x - 2x - 1x + 4 - 3("Is the sqrt(9)") - 2sqrt(3) + 2sqrt(3))
y=a(x^2 - 4x + 1)
Plug in points
5=a(2^2 - 4(2) + 1)
5=a(4 - 8 + 1)
5=a(3)
a=-5/3
Plug that our equation (x^2 - 4x + 1)
y=(-5/3^2 - 4/1(-5/3) + 1(-5/3))
y=25/9 - 20/3 - 5/3
ANSWER:
y=5/3 - 20/3 - 5/3