Solve 2*sqrt{x^2-3x}+3=x

$$2 \cdot \sqrt { x ^ { 2 } - 3 x } + 3 = x$$

Solve for x

$$x=3$$

Solution Steps

Subtract 3 from both sides of the equation.

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

Square both sides of the equation.

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

Expand \left(2\sqrt{x^{2}-3x}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 4 by x^{2}-3x.

$$4x^{2}-12x=\left(x-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$4x^{2}-12x=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$4x^{2}-12x-x^{2}=-6x+9$$

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

$$3x^{2}-12x=-6x+9$$

Add 6x to both sides.

$$3x^{2}-12x+6x=9$$

Combine -12x and 6x to get -6x.

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

Subtract 9 from both sides.

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

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.

$$a+b=-2 ab=1\left(-3\right)=-3$$

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.

$$a=-3 b=1$$

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

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

Factor out x in x^{2}-3x.

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

Factor out common term x-3 by using distributive property.

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

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

$$x=3 x=-1$$

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

$$2\sqrt{3^{2}-3\times 3}+3=3$$

Simplify. The value x=3 satisfies the equation.

$$3=3$$

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

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

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

$$7=-1$$

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

$$x=3$$