find equation of quadratic function.?

Question

that passes through(-4,5) if its zeros are 2+ sqrt( 3) and 2- sqrt (3)

Answer 1

Start with figuring out the equation of a line that has roots of 2+sqrt(3) and 2-sqrt(3).

y = 0 when:
x = 2+sqrt(3) or x = 2-sqrt(3)

x - (2+sqrt(3)) = 0
x - (2-sqrt(3)) = 0
y = [ x - (2+sqrt(3) ][ x - (2-sqrt(3)) ]

Now multiply out using the FOIL method (First, Outer, Inner, Last)
y = x² - (2-sqrt(3))x - (2+sqrt(3))x + (2+sqrt(3))(2-sqrt(3))

Expand out the x terms:
y = x² - 2x + sqrt(3)x - 2x - sqrt(3)x + (2+sqrt(3))(2-sqrt(3))

Cancel the +sqrt(3)x -sqrt(3)x
y = x² - 2x - 2x + (2+sqrt(3))(2-sqrt(3))

Combine x terms:
y = x² - 4x + (2+sqrt(3))(2-sqrt(3))

Multiply out the last part, again using FOIL:
y = x² - 4x + (2² - 2sqrt(3) + 2sqrt(3) -sqrt(3)²)

Cancel the - 2sqrt(3) + 2sqrt(3):
y = x² - 4x + (2² - sqrt(3)²)

Replace 2² = 4 and sqrt(3)² = 3
y = x² - 4x + (4 - 3)

Simplify:
y = x² - 4x + 1

Okay, but before you celebrate, this function has to go through (-4, 5)

Realize this will still equal zero if we multiply it by a constant. So the following equation will still go through the same roots:
y = k(x² - 4x + 1)

Now plug in x = -4 and y = 5 and solve for k:
5 = k((-4)² - 4(-4) + 1)
5 = k(16 +16 + 1)
5 = k(33)
k = 5/33

So the final answer is:
y = 5/33[x² - 4x + 1]