Solve 4R^2-R^2=3

$$4 R ^ { 2 } - R ^ { 2 } = 3$$

Solve for R

$$R=1$$

$$R=-1$$

Steps Using Difference of Squares

Combine 4R^{2} and -R^{2} to get 3R^{2}.

$$3R^{2}=3$$

Subtract 3 from both sides.

$$3R^{2}-3=0$$

Divide both sides by 3.

$$R^{2}-1=0$$

Consider R^{2}-1. Rewrite R^{2}-1 as R^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(R-1\right)\left(R+1\right)=0$$

To find equation solutions, solve R-1=0 and R+1=0.

$$R=1 R=-1$$

Steps by Finding Square Root

Combine 4R^{2} and -R^{2} to get 3R^{2}.

$$3R^{2}=3$$

Divide both sides by 3.

$$R^{2}=\frac{3}{3}$$

Divide 3 by 3 to get 1.

$$R^{2}=1$$

Take the square root of both sides of the equation.

$$R=1 R=-1$$

Steps Using the Quadratic Formula

Combine 4R^{2} and -R^{2} to get 3R^{2}.

$$3R^{2}=3$$

Subtract 3 from both sides.

$$3R^{2}-3=0$$

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

$$R=\frac{0±\sqrt{0^{2}-4\times 3\left(-3\right)}}{2\times 3}$$

Square 0.

$$R=\frac{0±\sqrt{-4\times 3\left(-3\right)}}{2\times 3}$$

Multiply -4 times 3.

$$R=\frac{0±\sqrt{-12\left(-3\right)}}{2\times 3}$$

Multiply -12 times -3.

$$R=\frac{0±\sqrt{36}}{2\times 3}$$

Take the square root of 36.

$$R=\frac{0±6}{2\times 3}$$

Multiply 2 times 3.

$$R=\frac{0±6}{6}$$

Now solve the equation R=\frac{0±6}{6} when ± is plus. Divide 6 by 6.

$$R=1$$

Now solve the equation R=\frac{0±6}{6} when ± is minus. Divide -6 by 6.

$$R=-1$$

The equation is now solved.

$$R=1 R=-1$$