Solve sqrt{x^2-2x+10}+2x+1=0

$$\sqrt { x ^ { 2 } - 2 x + 10 } + 2 x + 1 = 0$$

Solve for x

$$x=-3$$

Solution Steps

Subtract 2x+1 from both sides of the equation.

$$\sqrt{x^{2}-2x+10}=-\left(2x+1\right)$$

To find the opposite of 2x+1, find the opposite of each term.

$$\sqrt{x^{2}-2x+10}=-2x-1$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}-2x+10}\right)^{2}=\left(-2x-1\right)^{2}$$

Calculate \sqrt{x^{2}-2x+10} to the power of 2 and get x^{2}-2x+10.

$$x^{2}-2x+10=\left(-2x-1\right)^{2}$$

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

$$x^{2}-2x+10=4x^{2}+4x+1$$

Subtract 4x^{2} from both sides.

$$x^{2}-2x+10-4x^{2}=4x+1$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-2x+10=4x+1$$

Subtract 4x from both sides.

$$-3x^{2}-2x+10-4x=1$$

Combine -2x and -4x to get -6x.

$$-3x^{2}-6x+10=1$$

Subtract 1 from both sides.

$$-3x^{2}-6x+10-1=0$$

Subtract 1 from 10 to get 9.

$$-3x^{2}-6x+9=0$$

Divide both sides by 3.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3=-3$$

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. The only such pair is the system solution.

$$a=1 b=-3$$

Rewrite -x^{2}-2x+3 as \left(-x^{2}+x\right)+\left(-3x+3\right).

$$\left(-x^{2}+x\right)+\left(-3x+3\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(-x+1\right)+3\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+3\right)$$

To find equation solutions, solve -x+1=0 and x+3=0.

$$x=1 x=-3$$

Substitute 1 for x in the equation \sqrt{x^{2}-2x+10}+2x+1=0.

$$\sqrt{1^{2}-2+10}+2\times 1+1=0$$

Simplify. The value x=1 does not satisfy the equation.

$$6=0$$

Substitute -3 for x in the equation \sqrt{x^{2}-2x+10}+2x+1=0.

$$\sqrt{\left(-3\right)^{2}-2\left(-3\right)+10}+2\left(-3\right)+1=0$$

Simplify. The value x=-3 satisfies the equation.

$$0=0$$

Equation \sqrt{x^{2}-2x+10}=-2x-1 has a unique solution.

$$x=-3$$