Solve sqrt{16+2x}*-4=2x

$$\sqrt{ 16+2x } \times -4=2x$$

Solve for x

$$x=4-4\sqrt{5}\approx -4.94427191$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{16+2x}\left(-4\right)\right)^{2}=\left(2x\right)^{2}$$

Expand \left(\sqrt{16+2x}\left(-4\right)\right)^{2}.

$$\left(\sqrt{16+2x}\right)^{2}\left(-4\right)^{2}=\left(2x\right)^{2}$$

Calculate \sqrt{16+2x} to the power of 2 and get 16+2x.

$$\left(16+2x\right)\left(-4\right)^{2}=\left(2x\right)^{2}$$

Calculate -4 to the power of 2 and get 16.

$$\left(16+2x\right)\times 16=\left(2x\right)^{2}$$

Use the distributive property to multiply 16+2x by 16.

$$256+32x=\left(2x\right)^{2}$$

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

$$256+32x=2^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$256+32x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$256+32x-4x^{2}=0$$

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.

$$-4x^{2}+32x+256=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 32 for b, and 256 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}-4\left(-4\right)\times 256}}{2\left(-4\right)}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\left(-4\right)\times 256}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-32±\sqrt{1024+16\times 256}}{2\left(-4\right)}$$

Multiply 16 times 256.

$$x=\frac{-32±\sqrt{1024+4096}}{2\left(-4\right)}$$

Add 1024 to 4096.

$$x=\frac{-32±\sqrt{5120}}{2\left(-4\right)}$$

Take the square root of 5120.

$$x=\frac{-32±32\sqrt{5}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-32±32\sqrt{5}}{-8}$$

Now solve the equation x=\frac{-32±32\sqrt{5}}{-8} when ± is plus. Add -32 to 32\sqrt{5}.

$$x=\frac{32\sqrt{5}-32}{-8}$$

Divide -32+32\sqrt{5} by -8.

$$x=4-4\sqrt{5}$$

Now solve the equation x=\frac{-32±32\sqrt{5}}{-8} when ± is minus. Subtract 32\sqrt{5} from -32.

$$x=\frac{-32\sqrt{5}-32}{-8}$$

Divide -32-32\sqrt{5} by -8.

$$x=4\sqrt{5}+4$$

The equation is now solved.

$$x=4-4\sqrt{5} x=4\sqrt{5}+4$$

Substitute 4-4\sqrt{5} for x in the equation \sqrt{16+2x}\left(-4\right)=2x.

$$\sqrt{16+2\left(4-4\sqrt{5}\right)}\left(-4\right)=2\left(4-4\sqrt{5}\right)$$

Simplify. The value x=4-4\sqrt{5} satisfies the equation.

$$-8\times 5^{\frac{1}{2}}+8=8-8\times 5^{\frac{1}{2}}$$

Substitute 4\sqrt{5}+4 for x in the equation \sqrt{16+2x}\left(-4\right)=2x.

$$\sqrt{16+2\left(4\sqrt{5}+4\right)}\left(-4\right)=2\left(4\sqrt{5}+4\right)$$

Simplify. The value x=4\sqrt{5}+4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-8\times 5^{\frac{1}{2}}-8=8\times 5^{\frac{1}{2}}+8$$

Equation -4\sqrt{2x+16}=2x has a unique solution.

$$x=4-4\sqrt{5}$$