Solve R^2=3^2+4^2

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

Solve for R

$$R=-5$$

$$R=5$$

Steps Using Difference of Squares

Calculate 3 to the power of 2 and get 9.

$$R^{2}=9+4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$R^{2}=9+16$$

Add 9 and 16 to get 25.

$$R^{2}=25$$

Subtract 25 from both sides.

$$R^{2}-25=0$$

Consider R^{2}-25. Rewrite R^{2}-25 as R^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(R-5\right)\left(R+5\right)=0$$

To find equation solutions, solve R-5=0 and R+5=0.

$$R=5 R=-5$$

Steps by Finding Square Root

Calculate 3 to the power of 2 and get 9.

$$R^{2}=9+4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$R^{2}=9+16$$

Add 9 and 16 to get 25.

$$R^{2}=25$$

Take the square root of both sides of the equation.

$$R=5 R=-5$$

Steps Using the Quadratic Formula

Calculate 3 to the power of 2 and get 9.

$$R^{2}=9+4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$R^{2}=9+16$$

Add 9 and 16 to get 25.

$$R^{2}=25$$

Subtract 25 from both sides.

$$R^{2}-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$R=\frac{0±\sqrt{0^{2}-4\left(-25\right)}}{2}$$

Square 0.

$$R=\frac{0±\sqrt{-4\left(-25\right)}}{2}$$

Multiply -4 times -25.

$$R=\frac{0±\sqrt{100}}{2}$$

Take the square root of 100.

$$R=\frac{0±10}{2}$$

Now solve the equation R=\frac{0±10}{2} when ± is plus. Divide 10 by 2.

$$R=5$$

Now solve the equation R=\frac{0±10}{2} when ± is minus. Divide -10 by 2.

$$R=-5$$

The equation is now solved.

$$R=5 R=-5$$