Solve 2x-5=2sqrt{3-x}

$$2 x - 5 = 2 \sqrt { 3 - x }$$

Solve for x

$$x = \frac{\sqrt{3} + 4}{2} \approx 2.866025404$$

Solution Steps

Square both sides of the equation.

$$\left(2x-5\right)^{2}=\left(2\sqrt{3-x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$4x^{2}-20x+25=\left(2\sqrt{3-x}\right)^{2}$$

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

$$4x^{2}-20x+25=2^{2}\left(\sqrt{3-x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-20x+25=4\left(\sqrt{3-x}\right)^{2}$$

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

$$4x^{2}-20x+25=4\left(3-x\right)$$

Use the distributive property to multiply 4 by 3-x.

$$4x^{2}-20x+25=12-4x$$

Subtract 12 from both sides.

$$4x^{2}-20x+25-12=-4x$$

Subtract 12 from 25 to get 13.

$$4x^{2}-20x+13=-4x$$

Add 4x to both sides.

$$4x^{2}-20x+13+4x=0$$

Combine -20x and 4x to get -16x.

$$4x^{2}-16x+13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -16 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 4\times 13}}{2\times 4}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 4\times 13}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-16\right)±\sqrt{256-16\times 13}}{2\times 4}$$

Multiply -16 times 13.

$$x=\frac{-\left(-16\right)±\sqrt{256-208}}{2\times 4}$$

Add 256 to -208.

$$x=\frac{-\left(-16\right)±\sqrt{48}}{2\times 4}$$

Take the square root of 48.

$$x=\frac{-\left(-16\right)±4\sqrt{3}}{2\times 4}$$

The opposite of -16 is 16.

$$x=\frac{16±4\sqrt{3}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{16±4\sqrt{3}}{8}$$

Now solve the equation x=\frac{16±4\sqrt{3}}{8} when ± is plus. Add 16 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+16}{8}$$

Divide 16+4\sqrt{3} by 8.

$$x=\frac{\sqrt{3}}{2}+2$$

Now solve the equation x=\frac{16±4\sqrt{3}}{8} when ± is minus. Subtract 4\sqrt{3} from 16.

$$x=\frac{16-4\sqrt{3}}{8}$$

Divide 16-4\sqrt{3} by 8.

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

The equation is now solved.

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

Substitute \frac{\sqrt{3}}{2}+2 for x in the equation 2x-5=2\sqrt{3-x}.

$$2\left(\frac{\sqrt{3}}{2}+2\right)-5=2\sqrt{3-\left(\frac{\sqrt{3}}{2}+2\right)}$$

Simplify. The value x=\frac{\sqrt{3}}{2}+2 satisfies the equation.

$$3^{\frac{1}{2}}-1=-1+3^{\frac{1}{2}}$$

Substitute -\frac{\sqrt{3}}{2}+2 for x in the equation 2x-5=2\sqrt{3-x}.

$$2\left(-\frac{\sqrt{3}}{2}+2\right)-5=2\sqrt{3-\left(-\frac{\sqrt{3}}{2}+2\right)}$$

Simplify. The value x=-\frac{\sqrt{3}}{2}+2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-3^{\frac{1}{2}}-1=1+3^{\frac{1}{2}}$$

Equation 2x-5=2\sqrt{3-x} has a unique solution.

$$x=\frac{\sqrt{3}}{2}+2$$