x^2-13=sqrt(x+13)
Assume that both sides of the above equation are positive numbers.
Find all possible values of x.
I think the "
trick" is "solving" for 13 rather than solving for a variable.
Pretend the integer 13 is a variable.
If each side of the expression equals K=LHS=RHS, then you get this system of equations.
13 = x^2 - K
13 = K^2 - x
subtracting:
0 = (x^2 - K^2) + (x - K)
0 = (x-K)*(x+K) + 1*(x - K)
0 = (x-K)*(x+K+1)
x = either K or -(K+1)
case 1: x=K, 13 = x^2 - K
13 = x^2 - x; x^2 - x - 13 = 0
x = (1 + sqrt(53))/2 or (1 - sqrt(53))/2 ;
but eliminate the negative option
case2: x = -(K+1), 13 = x^2 - K
k = -x - 1
x^2 + x - 12 = 0
x = -(1/2) ± (7/2) = (-4, 3); but eliminate the 3
Note that since in case 1, the assumption is x=K, the first value of x ie (1 + sqrt(53))/2 is also equal to the LHS and the RHS of the original equation.
2 of the 4 potential solutions because they made K be negative.
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Posted by Larry
on 2023-06-28 19:39:08 |
tricky? solution
