Solve x+1=sqrt{19-x}

$$x + 1 = \sqrt { 19 - x }$$

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(x+1\right)^{2}=\left(\sqrt{19-x}\right)^{2}$$

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

$$x^{2}+2x+1=\left(\sqrt{19-x}\right)^{2}$$

Calculate \sqrt{19-x} to the power of 2 and get 19-x.

$$x^{2}+2x+1=19-x$$

Subtract 19 from both sides.

$$x^{2}+2x+1-19=-x$$

Subtract 19 from 1 to get -18.

$$x^{2}+2x-18=-x$$

Add x to both sides.

$$x^{2}+2x-18+x=0$$

Combine 2x and x to get 3x.

$$x^{2}+3x-18=0$$

To solve the equation, factor x^{2}+3x-18 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.

$$a+b=3 ab=-18$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 3.

$$a=-3 b=6$$

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

$$\left(x-3\right)\left(x+6\right)$$

To find equation solutions, solve x-3=0 and x+6=0.

$$x=3 x=-6$$

Substitute 3 for x in the equation x+1=\sqrt{19-x}.

$$3+1=\sqrt{19-3}$$

Simplify. The value x=3 satisfies the equation.

$$4=4$$

Substitute -6 for x in the equation x+1=\sqrt{19-x}.

$$-6+1=\sqrt{19-\left(-6\right)}$$

Simplify. The value x=-6 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-5=5$$

Equation x+1=\sqrt{19-x} has a unique solution.

$$x=3$$