Solve sqrt{3x+4}=x^

$$\sqrt{ 3x+4 } = { x }^{ }$$

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x+4}\right)^{2}=\left(x^{1}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 1 and 2 to get 2.

$$\left(\sqrt{3x+4}\right)^{2}=x^{2}$$

Calculate \sqrt{3x+4} to the power of 2 and get 3x+4.

$$3x+4=x^{2}$$

Subtract x^{2} from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

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

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 -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 -x^{2}+3x+4 as \left(-x^{2}+4x\right)+\left(-x+4\right).

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

Factor out -x in the first and -1 in the second group.

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

Factor out common term x-4 by using distributive property.

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

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

$$x=4 x=-1$$

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

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

Simplify. The value x=4 satisfies the equation.

$$4=4$$

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

$$\sqrt{3\left(-1\right)+4}=\left(-1\right)^{1}$$

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

$$1=-1$$

Equation \sqrt{3x+4}=x^{1} has a unique solution.

$$x=4$$