Solve sqrt{x^2-9x+4}=sqrt{x^2-5}

$$\sqrt{ { x }^{ 2 } -9x+4 } = \sqrt{ { x }^{ 2 } -5 }$$

Solve for x (complex solution)

$$x=1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}-9x+4}\right)^{2}=\left(\sqrt{x^{2}-5}\right)^{2}$$

Calculate \sqrt{x^{2}-9x+4} to the power of 2 and get x^{2}-9x+4.

$$x^{2}-9x+4=\left(\sqrt{x^{2}-5}\right)^{2}$$

Calculate \sqrt{x^{2}-5} to the power of 2 and get x^{2}-5.

$$x^{2}-9x+4=x^{2}-5$$

Subtract x^{2} from both sides.

$$x^{2}-9x+4-x^{2}=-5$$

Combine x^{2} and -x^{2} to get 0.

$$-9x+4=-5$$

Subtract 4 from both sides.

$$-9x=-5-4$$

Subtract 4 from -5 to get -9.

$$-9x=-9$$

Divide both sides by -9.

$$x=\frac{-9}{-9}$$

Divide -9 by -9 to get 1.

$$x=1$$

Substitute 1 for x in the equation \sqrt{x^{2}-9x+4}=\sqrt{x^{2}-5}.

$$\sqrt{1^{2}-9+4}=\sqrt{1^{2}-5}$$

Simplify. The value x=1 satisfies the equation.

$$2i=2i$$

Equation \sqrt{x^{2}-9x+4}=\sqrt{x^{2}-5} has a unique solution.

$$x=1$$