a.
(k + 9)^2 = 71
(k + 9)(k + 9) = 71
k^2 + 9k + 9k + 81 = 71
k^2 + 18k + 81 = 71
k^2 + 18k + 81 - 71 = 0
k^2 + 18k + 10 = 0
Using the quadratic formula:
k = (-b ± √(b^2 - 4ac)) / 2a
k = (-18 ± √(18^2 - 4(1)(10))) / 2(1)
k = (-18 ± √(324 - 40)) / 2
k = (-18 ± √284) / 2
k = (-18 ± 16.86) / 2
k ≈ (-18 + 16.86) / 2 or k ≈ (-18 - 16.86) / 2
k ≈ -0.57 or k ≈ -17.43
b.
(6 - p)^2 = 47
(6 - p)(6 - p) = 47
36 - 6p - 6p + p^2 = 47
36 - 12p + p^2 = 47
p^2 - 12p + 36 = 47
p^2 - 12p - 11 = 0
Using the quadratic formula:
p = (-b ± √(b^2 - 4ac)) / 2a
p = (-(-12) ± √((-12)^2 - 4(1)(-11))) / 2(1)
p = (12 ± √(144 + 44)) / 2
p = (12 ± √188) / 2
p ≈ (12 + 13.71) / 2 or p ≈ (12 - 13.71) / 2
p ≈ 12.85 or p ≈ -0.85
Determine the approximate solutions to each of the given equations. Show your work. a.
(k + 9)2
= 71
b. (6 - p)2
47
1 answer
1 answer