Solve sqrt{5n-1}-n=1

$$\sqrt { 5 n - 1 } - n = 1$$

Solve for n

$$n=2$$

$$n=1$$

Solution Steps

Subtract -n from both sides of the equation.

$$\sqrt{5n-1}=1+n$$

Square both sides of the equation.

$$\left(\sqrt{5n-1}\right)^{2}=\left(1+n\right)^{2}$$

Calculate \sqrt{5n-1} to the power of 2 and get 5n-1.

$$5n-1=\left(1+n\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+n\right)^{2}.

$$5n-1=1+2n+n^{2}$$

Subtract 1 from both sides.

$$5n-1-1=2n+n^{2}$$

Subtract 1 from -1 to get -2.

$$5n-2=2n+n^{2}$$

Subtract 2n from both sides.

$$5n-2-2n=n^{2}$$

Combine 5n and -2n to get 3n.

$$3n-2=n^{2}$$

Subtract n^{2} from both sides.

$$3n-2-n^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-n^{2}+3n-2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -n^{2}+an+bn-2. To find a and b, set up a system to be solved.

$$a+b=3 ab=-\left(-2\right)=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=2 b=1$$

Rewrite -n^{2}+3n-2 as \left(-n^{2}+2n\right)+\left(n-2\right).

$$\left(-n^{2}+2n\right)+\left(n-2\right)$$

Factor out -n in -n^{2}+2n.

$$-n\left(n-2\right)+n-2$$

Factor out common term n-2 by using distributive property.

$$\left(n-2\right)\left(-n+1\right)$$

To find equation solutions, solve n-2=0 and -n+1=0.

$$n=2 n=1$$

Substitute 2 for n in the equation \sqrt{5n-1}-n=1.

$$\sqrt{5\times 2-1}-2=1$$

Simplify. The value n=2 satisfies the equation.

$$1=1$$

Substitute 1 for n in the equation \sqrt{5n-1}-n=1.

$$\sqrt{5\times 1-1}-1=1$$

Simplify. The value n=1 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{5n-1}=n+1.

$$n=2 n=1$$