Solve (2r)^2-r^2=3

$$( 2 r ) ^ { 2 } - r ^ { 2 } = 3$$

Solve for r

$$r=-1$$

$$r=1$$

Steps Using Difference of Squares

Expand \left(2r\right)^{2}.

$$2^{2}r^{2}-r^{2}=3$$

Calculate 2 to the power of 2 and get 4.

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

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

Expand \left(2r\right)^{2}.

$$2^{2}r^{2}-r^{2}=3$$

Calculate 2 to the power of 2 and get 4.

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

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

Expand \left(2r\right)^{2}.

$$2^{2}r^{2}-r^{2}=3$$

Calculate 2 to the power of 2 and get 4.

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

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.

$$r=1 r=-1$$