solving a word problem using a quadratic equation with irrational roots

Question

h = 50-10t-5t²

Answer 1

Use the quadratic formula.

Ax^2 + Bx + C = 0

A = -5

B = -10

C = 50

(-B ± √(B^2 -4AC)) / (2A)

(-(-10) ± √((-10)^2 -4(-5)(50))) / (2×(-5))

(10 ± √(100 + 1,000)) /(-10)

(10 ± √(1,100)) / (-10)

(10 ± 10√(11)) / ( -10 )

-1 ± √11

X₁ = -1 + √11

X₂ = -1 -√11

NOTE that -(±) is still ±. Why? Because the signs change, but we can switch order of the values to become positive first, negative second again. Recall that both numbers still have the same absolute value.

Answer 2

Use the quadratic formula:

[10±√(100-4(50)(-5))] / 2(-5)

(10±√1100) / -10

(10±10√11) / -10

Simplify:

-1±√11

Hope this helps!