Solve A^2=5-1^3=

$$A ^ { 2 } = 5 - 1 ^ { 3 } =$$

Solve for A

$$A=2$$

$$A=-2$$

Steps Using Difference of Squares

Calculate 1 to the power of 3 and get 1.

$$A^{2}=5-1$$

Subtract 1 from 5 to get 4.

$$A^{2}=4$$

Subtract 4 from both sides.

$$A^{2}-4=0$$

Consider A^{2}-4. Rewrite A^{2}-4 as A^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(A-2\right)\left(A+2\right)=0$$

To find equation solutions, solve A-2=0 and A+2=0.

$$A=2 A=-2$$

Steps by Finding Square Root

Calculate 1 to the power of 3 and get 1.

$$A^{2}=5-1$$

Subtract 1 from 5 to get 4.

$$A^{2}=4$$

Take the square root of both sides of the equation.

$$A=2 A=-2$$

Steps Using the Quadratic Formula

Calculate 1 to the power of 3 and get 1.

$$A^{2}=5-1$$

Subtract 1 from 5 to get 4.

$$A^{2}=4$$

Subtract 4 from both sides.

$$A^{2}-4=0$$

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

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

Square 0.

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

Multiply -4 times -4.

$$A=\frac{0±\sqrt{16}}{2}$$

Take the square root of 16.

$$A=\frac{0±4}{2}$$

Now solve the equation A=\frac{0±4}{2} when ± is plus. Divide 4 by 2.

$$A=2$$

Now solve the equation A=\frac{0±4}{2} when ± is minus. Divide -4 by 2.

$$A=-2$$

The equation is now solved.

$$A=2 A=-2$$