Solve {l}{a+b=6}{a^2+b^2=6}

$$\left. \begin{array} { l } { a + b = 6 } \\ { a ^ { 2 } + b ^ { 2 } = 6 } \end{array} \right.$$

Solve for a, b

$$a=3+\sqrt{6}i\approx 3+2.449489743i\text{, }b=-\sqrt{6}i+3\approx 3-2.449489743i$$

$$a=-\sqrt{6}i+3\approx 3-2.449489743i\text{, }b=3+\sqrt{6}i\approx 3+2.449489743i$$

Steps Using Substitution and Quadratic Formula

Solve a+b=6 for a by isolating a on the left hand side of the equal sign.

$$a+b=6$$

Subtract b from both sides of the equation.

$$a=-b+6$$

Substitute -b+6 for a in the other equation, b^{2}+a^{2}=6.

$$b^{2}+\left(-b+6\right)^{2}=6$$

Square -b+6.

$$b^{2}+b^{2}-12b+36=6$$

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

$$2b^{2}-12b+36=6$$

Subtract 6 from both sides of the equation.

$$2b^{2}-12b+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 6\left(-1\right)\times 2 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 30}}{2\times 2}$$

Square 1\times 6\left(-1\right)\times 2.

$$b=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 30}}{2\times 2}$$

Multiply -4 times 1+1\left(-1\right)^{2}.

$$b=\frac{-\left(-12\right)±\sqrt{144-8\times 30}}{2\times 2}$$

Multiply -8 times 30.

$$b=\frac{-\left(-12\right)±\sqrt{144-240}}{2\times 2}$$

Add 144 to -240.

$$b=\frac{-\left(-12\right)±\sqrt{-96}}{2\times 2}$$

Take the square root of -96.

$$b=\frac{-\left(-12\right)±4\sqrt{6}i}{2\times 2}$$

The opposite of 1\times 6\left(-1\right)\times 2 is 12.

$$b=\frac{12±4\sqrt{6}i}{2\times 2}$$

Multiply 2 times 1+1\left(-1\right)^{2}.

$$b=\frac{12±4\sqrt{6}i}{4}$$

Now solve the equation b=\frac{12±4\sqrt{6}i}{4} when ± is plus. Add 12 to 4i\sqrt{6}.

$$b=\frac{12+4\sqrt{6}i}{4}$$

Divide 12+4i\sqrt{6} by 4.

$$b=3+\sqrt{6}i$$

Now solve the equation b=\frac{12±4\sqrt{6}i}{4} when ± is minus. Subtract 4i\sqrt{6} from 12.

$$b=\frac{-4\sqrt{6}i+12}{4}$$

Divide 12-4i\sqrt{6} by 4.

$$b=-\sqrt{6}i+3$$

There are two solutions for b: 3+i\sqrt{6} and 3-i\sqrt{6}. Substitute 3+i\sqrt{6} for b in the equation a=-b+6 to find the corresponding solution for a that satisfies both equations.

$$a=-\left(3+\sqrt{6}i\right)+6$$

Now substitute 3-i\sqrt{6} for b in the equation a=-b+6 and solve to find the corresponding solution for a that satisfies both equations.

$$a=-\left(-\sqrt{6}i+3\right)+6$$

The system is now solved.

$$a=-\left(3+\sqrt{6}i\right)+6,b=3+\sqrt{6}i\text{ or }a=-\left(-\sqrt{6}i+3\right)+6,b=-\sqrt{6}i+3$$