Solve x+1=sqrt{3x+7}

$$x + 1 = \sqrt { 3 x + 7 }$$

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(x+1\right)^{2}=\left(\sqrt{3x+7}\right)^{2}$$

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

$$x^{2}+2x+1=\left(\sqrt{3x+7}\right)^{2}$$

Calculate \sqrt{3x+7} to the power of 2 and get 3x+7.

$$x^{2}+2x+1=3x+7$$

Subtract 3x from both sides.

$$x^{2}+2x+1-3x=7$$

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

$$x^{2}-x+1=7$$

Subtract 7 from both sides.

$$x^{2}-x+1-7=0$$

Subtract 7 from 1 to get -6.

$$x^{2}-x-6=0$$

To solve the equation, factor x^{2}-x-6 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.

$$a+b=-1 ab=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

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

$$\left(x-3\right)\left(x+2\right)$$

To find equation solutions, solve x-3=0 and x+2=0.

$$x=3 x=-2$$

Substitute 3 for x in the equation x+1=\sqrt{3x+7}.

$$3+1=\sqrt{3\times 3+7}$$

Simplify. The value x=3 satisfies the equation.

$$4=4$$

Substitute -2 for x in the equation x+1=\sqrt{3x+7}.

$$-2+1=\sqrt{3\left(-2\right)+7}$$

Simplify. The value x=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

Equation x+1=\sqrt{3x+7} has a unique solution.

$$x=3$$