How do I solve this equation?

Question

n - 3*sqrt(n) = 28


How do I solve this?

Answer 1

It is a quadratic equation in sqrt(n).
Use the substitution

x^2 = n

x^2 - 3 x = 28
x^2 - 3 x - 28 = 0
(x - 7) (x + 4) = 0

x = 7 or x= -4

Check the validity of the roots:

x= 7
n=49
49 - 3 sqrt(49) =? 28
49 - 3 7 =? 28
49 -21 =? 28
28 = 28 check

x=-4
n=16
16 - 3sqrt(16) =? 28
16 - 3 4 =? 28
16 - 12 =? 28
4 is not equal to 28 (doesn't check).

The square root symbol indicates the positive or principal root, not the negative one.
x=-4 and n=16 is a superfluous solution.

The only valid solution is n=49.

Answer 2

Add 3*sqrt(n) and minus 28 from the equation
n - 3*sqrt(n) + 3*sqrt(n) - 28 = 28 - 28 + 3*sqrt(n)
n -28 = 3*sqrt(n)
square the equation
(n - 28)^2 = 9n
n^2 - 56n + 784 = 9n
Minus 9n to the equation
n^2 - 56n + 784 - 9n = 9n - 9n
n^2 - 65n + 784 = 0
(n - 49)(n - 16) = 0
n = 49 or 16
only 49 satisfied the equation

Answer 3

n-3*sqrt(n)=28
n-28=3*sqrt(n)
n/3-28/3=sqrt(n)
(n/3-28/3)^2=n
foil the left side of the equation, then subtract n from both sides and use the quadratic equation to solve.

Answer 4

n = 7

Answer 5

3*sqrt(n) = n-28
squaring both sides:
9 n = n^2 + 784 - 56n
n^2 - 65n + 784 = 0
Solve rest yourself by using the quadratic formula.

Answer 6

By using mathematics. It was developed thousands of years ago, and has shown great utility in solving problems of this type.

Me? I'd just realize it is a polynomial for sqrt(n).

Answer 7

n - 3sqrtn = 28
Isolate the radical
-3sqrt(n) = 28 - n
Square both sides
9n = (28 - n)^2
9n = n^2 - 56n + 784
0 = n^2 - 65n + 784
n = 49 or n = 16
Only 49 checks out.

Answer 8

9n = (n - 28)^2 = n^2 - 56n + 784
n^2 - 65n + 784 = 0
(n-49)(n-16) = 0

n = 49 or 16