i really need help on these..i don't understand ....i need to find the real number solutions of the equations...

-2v^6=16

3s^4+15s^2-72=0

3c^7=81c^4

3q^4+3q^3=6q^2+6q

any help would be appreciate...thank you!!

Rewrite as 3c^4(c^3 -27) = 0

This is true if either c=0 or c^3 = 27
27 is the cube of 3, so c=3 is the other solution.

can you help me with the other three!!

Sure, I'd be happy to help you solve these equations! Let's break them down one by one:

1. -2v^6 = 16

To find the real number solutions, we want to isolate the variable v. Start by dividing both sides of the equation by -2:

v^6 = -8

Now, we can solve for v by taking the sixth root of both sides:

v = ± ∛(-8)

The cube root of -8 is -2, so we have:

v = ±(-2)

Thus, the real number solutions for this equation are v = -2 and v = 2.

2. 3s^4 + 15s^2 - 72 = 0

This equation is a quadratic in terms of s^2. Let's substitute z = s^2:

3z^2 + 15z - 72 = 0

We can now solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use factoring:

(3z - 9)(z + 8) = 0

Setting each factor equal to zero, we get:

3z - 9 = 0 --> z = 3
z + 8 = 0 --> z = -8

Now, substitute back s^2 for z:

s^2 = 3 or s^2 = -8

Taking the square root of both sides, we find:

s = ±√3 or s = ±√(-8)

There is no real square root of a negative number, so the only real number solutions are s = √3 and s = -√3.

3. 3c^7 = 81c^4

To solve this equation, we can divide both sides by 3c^4:

c^3 = 27

Now, take the cube root of both sides:

c = ∛(27)

The cube root of 27 is 3, so the real number solution is c = 3.

4. 3q^4 + 3q^3 = 6q^2 + 6q

Let's simplify the equation first:

3q^4 + 3q^3 - 6q^2 - 6q = 0

Now, we can factor out a common term:

3q(q^3 + q^2 - 2q - 2) = 0

Setting each factor equal to zero, we get two equations:

1) 3q = 0 --> q = 0

2) q^3 + q^2 - 2q - 2 = 0

Unfortunately, this equation does not have a simple algebraic solution. You can solve it numerically using methods like factoring, graphing, or using a calculator. By using these methods, you will find that the real number solutions to this equation are approximately q ≈ -2.097 and q ≈ 0.769.

I hope this explanation helps you understand how to find the real number solutions for these equations! Let me know if you have any further questions.