Solve n-6=sqrt{2n-9}

$$n - 6 = \sqrt { 2 n - 9 }$$

Solve for n

$$n=9$$

Solution Steps

Square both sides of the equation.

$$\left(n-6\right)^{2}=\left(\sqrt{2n-9}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-6\right)^{2}.

$$n^{2}-12n+36=\left(\sqrt{2n-9}\right)^{2}$$

Calculate \sqrt{2n-9} to the power of 2 and get 2n-9.

$$n^{2}-12n+36=2n-9$$

Subtract 2n from both sides.

$$n^{2}-12n+36-2n=-9$$

Combine -12n and -2n to get -14n.

$$n^{2}-14n+36=-9$$

Add 9 to both sides.

$$n^{2}-14n+36+9=0$$

Add 36 and 9 to get 45.

$$n^{2}-14n+45=0$$

To solve the equation, factor n^{2}-14n+45 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.

$$a+b=-14 ab=45$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -14.

$$a=-9 b=-5$$

Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.

$$\left(n-9\right)\left(n-5\right)$$

To find equation solutions, solve n-9=0 and n-5=0.

$$n=9 n=5$$

Substitute 9 for n in the equation n-6=\sqrt{2n-9}.

$$9-6=\sqrt{2\times 9-9}$$

Simplify. The value n=9 satisfies the equation.

$$3=3$$

Substitute 5 for n in the equation n-6=\sqrt{2n-9}.

$$5-6=\sqrt{2\times 5-9}$$

Simplify. The value n=5 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

Equation n-6=\sqrt{2n-9} has a unique solution.

$$n=9$$