√(2t + 1) = 5 - √t
squaring ... 2t + 1 = t - 10√t + 25
subtracting ... t + 10√t - 24 = 0
let a = √t ... a^2 + 10a - 24 = 0
factoring ... (a + 12)(a - 2) = 0 ... a = -12 , a = 2 ... t = 144 , t = 4
substitute back to confirm solution(s)
I need help with this question. I need to know how to do the steps for it. (hope it makes sense)
Question:
Square root 2t+1 - 5 = - square root t
2 answers
I will read it as:
√(2t+1) - 5 = -√t
√(2t+1) = 5 - √t
square both sides:
2t+1 = 25 - 10√t + t
10√t = 24 - t
square again
100t = 576 - 48t + t^2
t^2 - 148t + 576 = 0
(t - 144)(t - 4) = 0
t= 144 or t = 4, BUT since we squared the equation each of
the answers must be verified in the original equation
if t = 144
LS = √(288+1) - 5 = 2
RS = -√144 = -12
≠ LS, so x = 144 does not work
if x = 5
LS = √9 - 5 = -2
RS = -2
so x = 4 is the only solution
√(2t+1) - 5 = -√t
√(2t+1) = 5 - √t
square both sides:
2t+1 = 25 - 10√t + t
10√t = 24 - t
square again
100t = 576 - 48t + t^2
t^2 - 148t + 576 = 0
(t - 144)(t - 4) = 0
t= 144 or t = 4, BUT since we squared the equation each of
the answers must be verified in the original equation
if t = 144
LS = √(288+1) - 5 = 2
RS = -√144 = -12
≠ LS, so x = 144 does not work
if x = 5
LS = √9 - 5 = -2
RS = -2
so x = 4 is the only solution