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.