Solve sqrt{4x-3}+2=x

$$\sqrt { 4 x - 3 } + 2 = x$$

Solve for x

$$x=7$$

Solution Steps

Subtract 2 from both sides of the equation.

$$\sqrt{4x-3}=x-2$$

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Add 4x to both sides.

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

Combine 4x and 4x to get 8x.

$$8x-3-x^{2}=4$$

Subtract 4 from both sides.

$$8x-3-x^{2}-4=0$$

Subtract 4 from -3 to get -7.

$$8x-7-x^{2}=0$$

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

$$-x^{2}+8x-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=8 ab=-\left(-7\right)=7$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=7 b=1$$

Rewrite -x^{2}+8x-7 as \left(-x^{2}+7x\right)+\left(x-7\right).

$$\left(-x^{2}+7x\right)+\left(x-7\right)$$

Factor out -x in -x^{2}+7x.

$$-x\left(x-7\right)+x-7$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(-x+1\right)$$

To find equation solutions, solve x-7=0 and -x+1=0.

$$x=7 x=1$$

Substitute 7 for x in the equation \sqrt{4x-3}+2=x.

$$\sqrt{4\times 7-3}+2=7$$

Simplify. The value x=7 satisfies the equation.

$$7=7$$

Substitute 1 for x in the equation \sqrt{4x-3}+2=x.

$$\sqrt{4\times 1-3}+2=1$$

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

$$3=1$$

Equation \sqrt{4x-3}=x-2 has a unique solution.

$$x=7$$