Solve 2t=sqrt{(4-4t^2)+(2t)^2}

$$2 t = \sqrt { ( 4 - 4 t ^ { 2 } ) + ( 2 t ) ^ { 2 } }$$

Solve for t

$$t=1$$

Solution Steps

Square both sides of the equation.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

Combine -4t^{2} and 4t^{2} to get 0.

$$4t^{2}=\left(\sqrt{4}\right)^{2}$$

The square of \sqrt{4} is 4.

$$4t^{2}=4$$

Divide both sides by 4.

$$t^{2}=\frac{4}{4}$$

Divide 4 by 4 to get 1.

$$t^{2}=1$$

Subtract 1 from both sides.

$$t^{2}-1=0$$

Consider t^{2}-1. Rewrite t^{2}-1 as t^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(t-1\right)\left(t+1\right)=0$$

To find equation solutions, solve t-1=0 and t+1=0.

$$t=1 t=-1$$

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

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

Simplify. The value t=1 satisfies the equation.

$$2=2$$

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

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

Simplify. The value t=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

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

$$t=1$$