Solve x-4=sqrt{2x-9}

$$x - 4 = \sqrt { 2 x - 9 }$$

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

$$\left(x-4\right)^{2}=\left(\sqrt{2x-9}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16=\left(\sqrt{2x-9}\right)^{2}$$

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

$$x^{2}-8x+16=2x-9$$

Subtract 2x from both sides.

$$x^{2}-8x+16-2x=-9$$

Combine -8x and -2x to get -10x.

$$x^{2}-10x+16=-9$$

Add 9 to both sides.

$$x^{2}-10x+16+9=0$$

Add 16 and 9 to get 25.

$$x^{2}-10x+25=0$$

To solve the equation, factor x^{2}-10x+25 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=25$$

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 25.

$$-1,-25 -5,-5$$

Calculate the sum for each pair.

$$-1-25=-26 -5-5=-10$$

The solution is the pair that gives sum -10.

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

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

$$\left(x-5\right)\left(x-5\right)$$

Rewrite as a binomial square.

$$\left(x-5\right)^{2}$$

To find equation solution, solve x-5=0.

$$x=5$$

Substitute 5 for x in the equation x-4=\sqrt{2x-9}.

$$5-4=\sqrt{2\times 5-9}$$

Simplify. The value x=5 satisfies the equation.

$$1=1$$

Equation x-4=\sqrt{2x-9} has a unique solution.

$$x=5$$