Solve x-1=sqrt{x+5}

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

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x from both sides.

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

Combine -2x and -x to get -3x.

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

Subtract 5 from both sides.

$$x^{2}-3x+1-5=0$$

Subtract 5 from 1 to get -4.

$$x^{2}-3x-4=0$$

To solve the equation, factor x^{2}-3x-4 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.

$$a+b=-3 ab=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

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

$$\left(x-4\right)\left(x+1\right)$$

To find equation solutions, solve x-4=0 and x+1=0.

$$x=4 x=-1$$

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

$$4-1=\sqrt{4+5}$$

Simplify. The value x=4 satisfies the equation.

$$3=3$$

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

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

Simplify. The value x=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

Equation x-1=\sqrt{x+5} has a unique solution.

$$x=4$$