Solve sqrt{6+x^2}=1-x

$$\sqrt{ 6+ { x }^{ 2 } } = 1-x$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{6+x^{2}}\right)^{2}=\left(1-x\right)^{2}$$

Calculate \sqrt{6+x^{2}} to the power of 2 and get 6+x^{2}.

$$6+x^{2}=\left(1-x\right)^{2}$$

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

$$6+x^{2}=1-2x+x^{2}$$

Add 2x to both sides.

$$6+x^{2}+2x=1+x^{2}$$

Subtract x^{2} from both sides.

$$6+x^{2}+2x-x^{2}=1$$

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

$$6+2x=1$$

Subtract 6 from both sides.

$$2x=1-6$$

Subtract 6 from 1 to get -5.

$$2x=-5$$

Divide both sides by 2.

$$x=\frac{-5}{2}$$

Fraction \frac{-5}{2} can be rewritten as -\frac{5}{2} by extracting the negative sign.

$$x=-\frac{5}{2}$$

Substitute -\frac{5}{2} for x in the equation \sqrt{6+x^{2}}=1-x.

$$\sqrt{6+\left(-\frac{5}{2}\right)^{2}}=1-\left(-\frac{5}{2}\right)$$

Simplify. The value x=-\frac{5}{2} satisfies the equation.

$$\frac{7}{2}=\frac{7}{2}$$

Equation \sqrt{x^{2}+6}=1-x has a unique solution.

$$x=-\frac{5}{2}$$