Solve 3p^2+p^2=8^2

$$3 { p }^{ 2 } + { p }^{ 2 } = { 8 }^{ 2 }$$

Solve for p

$$p=4$$

$$p=-4$$

Steps Using Difference of Squares

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

$$4p^{2}=8^{2}$$

Calculate 8 to the power of 2 and get 64.

$$4p^{2}=64$$

Subtract 64 from both sides.

$$4p^{2}-64=0$$

Divide both sides by 4.

$$p^{2}-16=0$$

Consider p^{2}-16. Rewrite p^{2}-16 as p^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(p-4\right)\left(p+4\right)=0$$

To find equation solutions, solve p-4=0 and p+4=0.

$$p=4 p=-4$$

Steps by Finding Square Root

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

$$4p^{2}=8^{2}$$

Calculate 8 to the power of 2 and get 64.

$$4p^{2}=64$$

Divide both sides by 4.

$$p^{2}=\frac{64}{4}$$

Divide 64 by 4 to get 16.

$$p^{2}=16$$

Take the square root of both sides of the equation.

$$p=4 p=-4$$

Steps Using the Quadratic Formula

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

$$4p^{2}=8^{2}$$

Calculate 8 to the power of 2 and get 64.

$$4p^{2}=64$$

Subtract 64 from both sides.

$$4p^{2}-64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 0 for b, and -64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{0±\sqrt{0^{2}-4\times 4\left(-64\right)}}{2\times 4}$$

Square 0.

$$p=\frac{0±\sqrt{-4\times 4\left(-64\right)}}{2\times 4}$$

Multiply -4 times 4.

$$p=\frac{0±\sqrt{-16\left(-64\right)}}{2\times 4}$$

Multiply -16 times -64.

$$p=\frac{0±\sqrt{1024}}{2\times 4}$$

Take the square root of 1024.

$$p=\frac{0±32}{2\times 4}$$

Multiply 2 times 4.

$$p=\frac{0±32}{8}$$

Now solve the equation p=\frac{0±32}{8} when ± is plus. Divide 32 by 8.

$$p=4$$

Now solve the equation p=\frac{0±32}{8} when ± is minus. Divide -32 by 8.

$$p=-4$$

The equation is now solved.

$$p=4 p=-4$$