Solve sqrt{17+2x-3x^2}=x+1

$$\sqrt { 17 + 2 x - 3 x ^ { 2 } } = x + 1$$

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{17+2x-3x^{2}}\right)^{2}=\left(x+1\right)^{2}$$

Calculate \sqrt{17+2x-3x^{2}} to the power of 2 and get 17+2x-3x^{2}.

$$17+2x-3x^{2}=\left(x+1\right)^{2}$$

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

$$17+2x-3x^{2}=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$17+2x-3x^{2}-x^{2}=2x+1$$

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

$$17+2x-4x^{2}=2x+1$$

Subtract 2x from both sides.

$$17+2x-4x^{2}-2x=1$$

Combine 2x and -2x to get 0.

$$17-4x^{2}=1$$

Subtract 17 from both sides.

$$-4x^{2}=1-17$$

Subtract 17 from 1 to get -16.

$$-4x^{2}=-16$$

Divide both sides by -4.

$$x^{2}=\frac{-16}{-4}$$

Divide -16 by -4 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Substitute 2 for x in the equation \sqrt{17+2x-3x^{2}}=x+1.

$$\sqrt{17+2\times 2-3\times 2^{2}}=2+1$$

Simplify. The value x=2 satisfies the equation.

$$3=3$$

Substitute -2 for x in the equation \sqrt{17+2x-3x^{2}}=x+1.

$$\sqrt{17+2\left(-2\right)-3\left(-2\right)^{2}}=-2+1$$

Simplify. The value x=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-1$$

Equation \sqrt{17+2x-3x^{2}}=x+1 has a unique solution.

$$x=2$$