Solve sqrt{x^2-2^2}=sqrt{(x+0.5)^2-(1.5)^2}-1

$$\sqrt { x ^ { 2 } - 2 ^ { 2 } } = \sqrt { ( x + 0.5 ) ^ { 2 } - ( 1.5 ) ^ { 2 } } - 1$$

Solve for x

$$x = \frac{2 \sqrt{13} + 1}{3} \approx 2.737034184$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}-2^{2}}\right)^{2}=\left(\sqrt{\left(x+0.5\right)^{2}-1.5^{2}}-1\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$\left(\sqrt{x^{2}-4}\right)^{2}=\left(\sqrt{\left(x+0.5\right)^{2}-1.5^{2}}-1\right)^{2}$$

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

$$x^{2}-4=\left(\sqrt{\left(x+0.5\right)^{2}-1.5^{2}}-1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+0.5\right)^{2}.

$$x^{2}-4=\left(\sqrt{x^{2}+x+0.25-1.5^{2}}-1\right)^{2}$$

Calculate 1.5 to the power of 2 and get 2.25.

$$x^{2}-4=\left(\sqrt{x^{2}+x+0.25-2.25}-1\right)^{2}$$

Subtract 2.25 from 0.25 to get -2.

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

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

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

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

$$x^{2}-4=x^{2}+x-2-2\sqrt{x^{2}+x-2}+1$$

Add -2 and 1 to get -1.

$$x^{2}-4=x^{2}+x-1-2\sqrt{x^{2}+x-2}$$

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

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

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

$$x^{2}-4-x^{2}-x+1=-2\sqrt{x^{2}+x-2}$$

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

$$-4-x+1=-2\sqrt{x^{2}+x-2}$$

Add -4 and 1 to get -3.

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

Square both sides of the equation.

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

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

$$9+6x+x^{2}=\left(-2\sqrt{x^{2}+x-2}\right)^{2}$$

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

$$9+6x+x^{2}=\left(-2\right)^{2}\left(\sqrt{x^{2}+x-2}\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$9+6x+x^{2}=4\left(\sqrt{x^{2}+x-2}\right)^{2}$$

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

$$9+6x+x^{2}=4\left(x^{2}+x-2\right)$$

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

$$9+6x+x^{2}=4x^{2}+4x-8$$

Subtract 4x^{2} from both sides.

$$9+6x+x^{2}-4x^{2}=4x-8$$

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

$$9+6x-3x^{2}=4x-8$$

Subtract 4x from both sides.

$$9+6x-3x^{2}-4x=-8$$

Combine 6x and -4x to get 2x.

$$9+2x-3x^{2}=-8$$

Add 8 to both sides.

$$9+2x-3x^{2}+8=0$$

Add 9 and 8 to get 17.

$$17+2x-3x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-3x^{2}+2x+17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 2 for b, and 17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-3\right)\times 17}}{2\left(-3\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-3\right)\times 17}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-2±\sqrt{4+12\times 17}}{2\left(-3\right)}$$

Multiply 12 times 17.

$$x=\frac{-2±\sqrt{4+204}}{2\left(-3\right)}$$

Add 4 to 204.

$$x=\frac{-2±\sqrt{208}}{2\left(-3\right)}$$

Take the square root of 208.

$$x=\frac{-2±4\sqrt{13}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-2±4\sqrt{13}}{-6}$$

Now solve the equation x=\frac{-2±4\sqrt{13}}{-6} when ± is plus. Add -2 to 4\sqrt{13}.

$$x=\frac{4\sqrt{13}-2}{-6}$$

Divide -2+4\sqrt{13} by -6.

$$x=\frac{1-2\sqrt{13}}{3}$$

Now solve the equation x=\frac{-2±4\sqrt{13}}{-6} when ± is minus. Subtract 4\sqrt{13} from -2.

$$x=\frac{-4\sqrt{13}-2}{-6}$$

Divide -2-4\sqrt{13} by -6.

$$x=\frac{2\sqrt{13}+1}{3}$$

The equation is now solved.

$$x=\frac{1-2\sqrt{13}}{3} x=\frac{2\sqrt{13}+1}{3}$$

Substitute \frac{1-2\sqrt{13}}{3} for x in the equation \sqrt{x^{2}-2^{2}}=\sqrt{\left(x+0.5\right)^{2}-1.5^{2}}-1.

$$\sqrt{\left(\frac{1-2\sqrt{13}}{3}\right)^{2}-2^{2}}=\sqrt{\left(\frac{1-2\sqrt{13}}{3}+0.5\right)^{2}-1.5^{2}}-1$$

Simplify. The value x=\frac{1-2\sqrt{13}}{3} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{3}\times 13^{\frac{1}{2}}-\frac{2}{3}=\frac{2}{3}-\frac{1}{3}\times 13^{\frac{1}{2}}$$

Substitute \frac{2\sqrt{13}+1}{3} for x in the equation \sqrt{x^{2}-2^{2}}=\sqrt{\left(x+0.5\right)^{2}-1.5^{2}}-1.

$$\sqrt{\left(\frac{2\sqrt{13}+1}{3}\right)^{2}-2^{2}}=\sqrt{\left(\frac{2\sqrt{13}+1}{3}+0.5\right)^{2}-1.5^{2}}-1$$

Simplify. The value x=\frac{2\sqrt{13}+1}{3} satisfies the equation.

$$\frac{1}{3}\times 13^{\frac{1}{2}}+\frac{2}{3}=\frac{2}{3}+\frac{1}{3}\times 13^{\frac{1}{2}}$$

Equation \sqrt{x^{2}-4}=\sqrt{\left(x+0.5\right)^{2}-2.25}-1 has a unique solution.

$$x=\frac{2\sqrt{13}+1}{3}$$