Solve sqrt{x^2-5x+4}+1=x-3

$$\sqrt { x ^ { 2 } - 5 x + 4 } + 1 = x - 3$$

Solve for x

$$x=4$$

Solution Steps

Subtract 1 from both sides of the equation.

$$\sqrt{x^{2}-5x+4}=x-3-1$$

Subtract 1 from -3 to get -4.

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

Square both sides of the equation.

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

Calculate \sqrt{x^{2}-5x+4} to the power of 2 and get x^{2}-5x+4.

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

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

$$x^{2}-5x+4=x^{2}-8x+16$$

Subtract x^{2} from both sides.

$$x^{2}-5x+4-x^{2}=-8x+16$$

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

$$-5x+4=-8x+16$$

Add 8x to both sides.

$$-5x+4+8x=16$$

Combine -5x and 8x to get 3x.

$$3x+4=16$$

Subtract 4 from both sides.

$$3x=16-4$$

Subtract 4 from 16 to get 12.

$$3x=12$$

Divide both sides by 3.

$$x=\frac{12}{3}$$

Divide 12 by 3 to get 4.

$$x=4$$

Substitute 4 for x in the equation \sqrt{x^{2}-5x+4}+1=x-3.

$$\sqrt{4^{2}-5\times 4+4}+1=4-3$$

Simplify. The value x=4 satisfies the equation.

$$1=1$$

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

$$x=4$$