Solve {l}{C=sqrt{8}+sqrt{8}}{D=sqrt{8}-sqrt{8}}{a=C*D}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { C = \sqrt{8} + \sqrt{8} }\\ { D = \sqrt{8} - \sqrt{8} }\\ { a = C \cdot D }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for C, D, a, b

$$b=0$$

Solution Steps

Consider the first equation. Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$C=2\sqrt{2}+\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$C=2\sqrt{2}+2\sqrt{2}$$

Combine 2\sqrt{2} and 2\sqrt{2} to get 4\sqrt{2}.

$$C=4\sqrt{2}$$

Consider the second equation. Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$D=2\sqrt{2}-\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$D=2\sqrt{2}-2\sqrt{2}$$

Combine 2\sqrt{2} and -2\sqrt{2} to get 0.

$$D=0$$

Consider the third equation. Insert the known values of variables into the equation.

$$a=4\sqrt{2}\times 0$$

Multiply 4 and 0 to get 0.

$$a=0\sqrt{2}$$

Anything times zero gives zero.

$$a=0$$

Consider the fourth equation. Insert the known values of variables into the equation.

$$b=0$$

The system is now solved.

$$C=4\sqrt{2} D=0 a=0 b=0$$