Solve sqrt{x^2-3x+2}-sqrt{x^2+x+x}=1

$$\sqrt{ { x }^{ 2 } -3x+2 } - \sqrt{ { x }^{ 2 } +x+x } =1$$

Solve for x

$$x=-\frac{2\sqrt{15}}{21}+\frac{3}{7}\approx 0.059715872$$

Solution Steps

Subtract -\sqrt{x^{2}+x+x} from both sides of the equation.

$$\sqrt{x^{2}-3x+2}=1+\sqrt{x^{2}+x+x}$$

Combine x and x to get 2x.

$$\sqrt{x^{2}-3x+2}=1+\sqrt{x^{2}+2x}$$

Square both sides of the equation.

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

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

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

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

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

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

$$x^{2}-3x+2=1+2\sqrt{x^{2}+2x}+x^{2}+2x$$

Subtract 1+x^{2}+2x from both sides of the equation.

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

To find the opposite of 1+x^{2}+2x, find the opposite of each term.

$$x^{2}-3x+2-1-x^{2}-2x=2\sqrt{x^{2}+2x}$$

Subtract 1 from 2 to get 1.

$$x^{2}-3x+1-x^{2}-2x=2\sqrt{x^{2}+2x}$$

Combine x^{2} and -x^{2} to get 0.

$$-3x+1-2x=2\sqrt{x^{2}+2x}$$

Combine -3x and -2x to get -5x.

$$-5x+1=2\sqrt{x^{2}+2x}$$

Square both sides of the equation.

$$\left(-5x+1\right)^{2}=\left(2\sqrt{x^{2}+2x}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-5x+1\right)^{2}.

$$25x^{2}-10x+1=\left(2\sqrt{x^{2}+2x}\right)^{2}$$

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

$$25x^{2}-10x+1=2^{2}\left(\sqrt{x^{2}+2x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$25x^{2}-10x+1=4\left(\sqrt{x^{2}+2x}\right)^{2}$$

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

$$25x^{2}-10x+1=4\left(x^{2}+2x\right)$$

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

$$25x^{2}-10x+1=4x^{2}+8x$$

Subtract 4x^{2} from both sides.

$$25x^{2}-10x+1-4x^{2}=8x$$

Combine 25x^{2} and -4x^{2} to get 21x^{2}.

$$21x^{2}-10x+1=8x$$

Subtract 8x from both sides.

$$21x^{2}-10x+1-8x=0$$

Combine -10x and -8x to get -18x.

$$21x^{2}-18x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 21 for a, -18 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 21}}{2\times 21}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 21}}{2\times 21}$$

Multiply -4 times 21.

$$x=\frac{-\left(-18\right)±\sqrt{324-84}}{2\times 21}$$

Add 324 to -84.

$$x=\frac{-\left(-18\right)±\sqrt{240}}{2\times 21}$$

Take the square root of 240.

$$x=\frac{-\left(-18\right)±4\sqrt{15}}{2\times 21}$$

The opposite of -18 is 18.

$$x=\frac{18±4\sqrt{15}}{2\times 21}$$

Multiply 2 times 21.

$$x=\frac{18±4\sqrt{15}}{42}$$

Now solve the equation x=\frac{18±4\sqrt{15}}{42} when ± is plus. Add 18 to 4\sqrt{15}.

$$x=\frac{4\sqrt{15}+18}{42}$$

Divide 18+4\sqrt{15} by 42.

$$x=\frac{2\sqrt{15}}{21}+\frac{3}{7}$$

Now solve the equation x=\frac{18±4\sqrt{15}}{42} when ± is minus. Subtract 4\sqrt{15} from 18.

$$x=\frac{18-4\sqrt{15}}{42}$$

Divide 18-4\sqrt{15} by 42.

$$x=-\frac{2\sqrt{15}}{21}+\frac{3}{7}$$

The equation is now solved.

$$x=\frac{2\sqrt{15}}{21}+\frac{3}{7} x=-\frac{2\sqrt{15}}{21}+\frac{3}{7}$$

Substitute \frac{2\sqrt{15}}{21}+\frac{3}{7} for x in the equation \sqrt{x^{2}-3x+2}-\sqrt{x^{2}+x+x}=1.

$$\sqrt{\left(\frac{2\sqrt{15}}{21}+\frac{3}{7}\right)^{2}-3\left(\frac{2\sqrt{15}}{21}+\frac{3}{7}\right)+2}-\sqrt{\left(\frac{2\sqrt{15}}{21}+\frac{3}{7}\right)^{2}+\frac{2\sqrt{15}}{21}+\frac{3}{7}+\frac{2\sqrt{15}}{21}+\frac{3}{7}}=1$$

Simplify. The value x=\frac{2\sqrt{15}}{21}+\frac{3}{7} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

Substitute -\frac{2\sqrt{15}}{21}+\frac{3}{7} for x in the equation \sqrt{x^{2}-3x+2}-\sqrt{x^{2}+x+x}=1.

$$\sqrt{\left(-\frac{2\sqrt{15}}{21}+\frac{3}{7}\right)^{2}-3\left(-\frac{2\sqrt{15}}{21}+\frac{3}{7}\right)+2}-\sqrt{\left(-\frac{2\sqrt{15}}{21}+\frac{3}{7}\right)^{2}-\frac{2\sqrt{15}}{21}+\frac{3}{7}-\frac{2\sqrt{15}}{21}+\frac{3}{7}}=1$$

Simplify. The value x=-\frac{2\sqrt{15}}{21}+\frac{3}{7} satisfies the equation.

$$1=1$$

Equation \sqrt{x^{2}-3x+2}=\sqrt{x^{2}+2x}+1 has a unique solution.

$$x=-\frac{2\sqrt{15}}{21}+\frac{3}{7}$$