What is the answer to n(x-3)=sqrt(x-3) ?

The answer to n(x-3)=sqrt(x-3) is x=(3n^2+1)/(n^2) {n 0},x=3

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n(x-3)=sqrt(x-3)

n (x - 3) = x - 3

x = \frac{3 n ^{2} + 1}{n ^{2}} {n < 0 or n > 0}, x = 3

Solution steps

n (x - 3) = x - 3

Expand

n (x - 3) : n x - 3 n

n x - 3 n = x - 3

Square both sides:

n^{2} x^{2} - 6 n^{2} x + 9 n^{2} = x - 3

n^{2} x^{2} - 6 n^{2} x + 9 n^{2} = x - 3

Solve

n^{2} x^{2} - 6 n^{2} x + 9 n^{2} = x - 3 : x = \frac{3 n ^{2} + 1}{n ^{2}}, x = 3

x = \frac{3 n ^{2} + 1}{n ^{2}}, x = 3

Verify Solutions:

x = \frac{3 n ^{2} + 1}{n ^{2}} {n < 0 or n > 0}, x = 3

True

The solutions are

x = \frac{3 n ^{2} + 1}{n ^{2}} {n < 0 or n > 0}, x = 3

2 x - 5 - x - 2 = 2

13 + x = x^{2} + 1

2 + 3 x - 2 = x

\frac{1}{2 x} = 0

2 x + 6 = x + 2

What is the answer to n(x-3)=sqrt(x-3) ?
The answer to n(x-3)=sqrt(x-3) is x=(3n^2+1)/(n^2) {n<0\lor n>0},x=3