Solve {l}{c=3a}{c^2=a^2+4}

$$\left\{ \begin{array} { l } { c = 3 a } \\ { c ^ { 2 } = a ^ { 2 } + 4 } \end{array} \right.$$

Solve for c, a

$$c=-\frac{3\sqrt{2}}{2}\approx -2.121320344\text{, }a=-\frac{\sqrt{2}}{2}\approx -0.707106781$$

$$c=\frac{3\sqrt{2}}{2}\approx 2.121320344\text{, }a=\frac{\sqrt{2}}{2}\approx 0.707106781$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract 3a from both sides.

$$c-3a=0$$

Consider the second equation. Subtract a^{2} from both sides.

$$c^{2}-a^{2}=4$$

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.

$$c-3a=0,-a^{2}+c^{2}=4$$

Solve c-3a=0 for c by isolating c on the left hand side of the equal sign.

$$c-3a=0$$

Subtract -3a from both sides of the equation.

$$c=3a$$

Substitute 3a for c in the other equation, -a^{2}+c^{2}=4.

$$-a^{2}+\left(3a\right)^{2}=4$$

Square 3a.

$$-a^{2}+9a^{2}=4$$

Add -a^{2} to 9a^{2}.

$$8a^{2}=4$$

Subtract 4 from both sides of the equation.

$$8a^{2}-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+1\times 3^{2} for a, 1\times 0\times 2\times 3 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{0±\sqrt{0^{2}-4\times 8\left(-4\right)}}{2\times 8}$$

Square 1\times 0\times 2\times 3.

$$a=\frac{0±\sqrt{-4\times 8\left(-4\right)}}{2\times 8}$$

Multiply -4 times -1+1\times 3^{2}.

$$a=\frac{0±\sqrt{-32\left(-4\right)}}{2\times 8}$$

Multiply -32 times -4.

$$a=\frac{0±\sqrt{128}}{2\times 8}$$

Take the square root of 128.

$$a=\frac{0±8\sqrt{2}}{2\times 8}$$

Multiply 2 times -1+1\times 3^{2}.

$$a=\frac{0±8\sqrt{2}}{16}$$

Now solve the equation a=\frac{0±8\sqrt{2}}{16} when ± is plus.

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

Now solve the equation a=\frac{0±8\sqrt{2}}{16} when ± is minus.

$$a=-\frac{\sqrt{2}}{2}$$

There are two solutions for a: \frac{\sqrt{2}}{2} and -\frac{\sqrt{2}}{2}. Substitute \frac{\sqrt{2}}{2} for a in the equation c=3a to find the corresponding solution for c that satisfies both equations.

$$c=3\times \frac{\sqrt{2}}{2}$$

Now substitute -\frac{\sqrt{2}}{2} for a in the equation c=3a and solve to find the corresponding solution for c that satisfies both equations.

$$c=3\left(-\frac{\sqrt{2}}{2}\right)$$

The system is now solved.

$$c=3\times \frac{\sqrt{2}}{2},a=\frac{\sqrt{2}}{2}\text{ or }c=3\left(-\frac{\sqrt{2}}{2}\right),a=-\frac{\sqrt{2}}{2}$$