Solve x^2+3x^2=2sqrt5^2

$$x ^ { 2 } + 3 x ^ { 2 } = 2 \sqrt { 5 } ^ { 2 }$$

Solve for x

$$x = \frac{\sqrt{10}}{2} \approx 1.58113883$$

$$x = -\frac{\sqrt{10}}{2} \approx -1.58113883$$

Steps by Finding Square Root

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$4x^{2}=2\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$4x^{2}=2\times 5$$

Multiply 2 and 5 to get 10.

$$4x^{2}=10$$

Divide both sides by 4.

$$x^{2}=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}=\frac{5}{2}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{10}}{2} x=-\frac{\sqrt{10}}{2}$$

Steps Using the Quadratic Formula

Combine x^{2} and 3x^{2} to get 4x^{2}.

$$4x^{2}=2\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$4x^{2}=2\times 5$$

Multiply 2 and 5 to get 10.

$$4x^{2}=10$$

Subtract 10 from both sides.

$$4x^{2}-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 0 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 4\left(-10\right)}}{2\times 4}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 4\left(-10\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{0±\sqrt{-16\left(-10\right)}}{2\times 4}$$

Multiply -16 times -10.

$$x=\frac{0±\sqrt{160}}{2\times 4}$$

Take the square root of 160.

$$x=\frac{0±4\sqrt{10}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{0±4\sqrt{10}}{8}$$

Now solve the equation x=\frac{0±4\sqrt{10}}{8} when ± is plus.

$$x=\frac{\sqrt{10}}{2}$$

Now solve the equation x=\frac{0±4\sqrt{10}}{8} when ± is minus.

$$x=-\frac{\sqrt{10}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{10}}{2} x=-\frac{\sqrt{10}}{2}$$