Solve sqrt{-x^2+x+3}=sqrt{-(x-1)^2+4}

$$\sqrt { - x ^ { 2 } + x + 3 } = \sqrt { - ( x - 1 ) ^ { 2 } + 4 }$$

Solve for x

$$x=0$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{-x^{2}+x+3}\right)^{2}=\left(\sqrt{-\left(x-1\right)^{2}+4}\right)^{2}$$

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

$$-x^{2}+x+3=\left(\sqrt{-\left(x-1\right)^{2}+4}\right)^{2}$$

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

$$-x^{2}+x+3=\left(\sqrt{-\left(x^{2}-2x+1\right)+4}\right)^{2}$$

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

$$-x^{2}+x+3=\left(\sqrt{-x^{2}+2x-1+4}\right)^{2}$$

Add -1 and 4 to get 3.

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

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

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

Add x^{2} to both sides.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

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

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

$$-x=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since -1 is not equal to 0, x must be equal to 0.

$$x=0$$

Substitute 0 for x in the equation \sqrt{-x^{2}+x+3}=\sqrt{-\left(x-1\right)^{2}+4}.

$$\sqrt{-0^{2}+0+3}=\sqrt{-\left(0-1\right)^{2}+4}$$

Simplify. The value x=0 satisfies the equation.

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

Equation \sqrt{3+x-x^{2}}=\sqrt{-\left(x-1\right)^{2}+4} has a unique solution.

$$x=0$$