Solve {l}{a^2+b^2=2}{a=2b}

$$\left. \begin{array} { l } { a ^ { 2 } + b ^ { 2 } = 2 } \\ { a = 2 b } \end{array} \right.$$

Solve for a, b

$$a=-\frac{2\sqrt{10}}{5}\approx -1.264911064\text{, }b=-\frac{\sqrt{10}}{5}\approx -0.632455532$$

$$a=\frac{2\sqrt{10}}{5}\approx 1.264911064\text{, }b=\frac{\sqrt{10}}{5}\approx 0.632455532$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract 2b from both sides.

$$a-2b=0$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$a-2b=0,b^{2}+a^{2}=2$$

Solve a-2b=0 for a by isolating a on the left hand side of the equal sign.

$$a-2b=0$$

Subtract -2b from both sides of the equation.

$$a=2b$$

Substitute 2b for a in the other equation, b^{2}+a^{2}=2.

$$b^{2}+\left(2b\right)^{2}=2$$

Square 2b.

$$b^{2}+4b^{2}=2$$

Add b^{2} to 4b^{2}.

$$5b^{2}=2$$

Subtract 2 from both sides of the equation.

$$5b^{2}-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 2^{2} for a, 1\times 0\times 2\times 2 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\times 5\left(-2\right)}}{2\times 5}$$

Square 1\times 0\times 2\times 2.

$$b=\frac{0±\sqrt{-4\times 5\left(-2\right)}}{2\times 5}$$

Multiply -4 times 1+1\times 2^{2}.

$$b=\frac{0±\sqrt{-20\left(-2\right)}}{2\times 5}$$

Multiply -20 times -2.

$$b=\frac{0±\sqrt{40}}{2\times 5}$$

Take the square root of 40.

$$b=\frac{0±2\sqrt{10}}{2\times 5}$$

Multiply 2 times 1+1\times 2^{2}.

$$b=\frac{0±2\sqrt{10}}{10}$$

Now solve the equation b=\frac{0±2\sqrt{10}}{10} when ± is plus.

$$b=\frac{\sqrt{10}}{5}$$

Now solve the equation b=\frac{0±2\sqrt{10}}{10} when ± is minus.

$$b=-\frac{\sqrt{10}}{5}$$

There are two solutions for b: \frac{\sqrt{10}}{5} and -\frac{\sqrt{10}}{5}. Substitute \frac{\sqrt{10}}{5} for b in the equation a=2b to find the corresponding solution for a that satisfies both equations.

$$a=2\times \frac{\sqrt{10}}{5}$$

Now substitute -\frac{\sqrt{10}}{5} for b in the equation a=2b and solve to find the corresponding solution for a that satisfies both equations.

$$a=2\left(-\frac{\sqrt{10}}{5}\right)$$

The system is now solved.

$$a=2\times \frac{\sqrt{10}}{5},b=\frac{\sqrt{10}}{5}\text{ or }a=2\left(-\frac{\sqrt{10}}{5}\right),b=-\frac{\sqrt{10}}{5}$$