Find the real-number solution of the equation. 3x^7 = 81x^4
I got this far...
3x^7 - 81x^4 = 0
x^4 (3x^3 - 81) = 0
3x^7 = 81x^4
divided both sides by 3
x^7 = 27x^4
divided both sides by x^4
x^3 = 27
take the cube root of each side
x=3
or
x = 0
since ....
x&7 - 27x^4 = 0
x^4 (x^3 - 27) = 0
x^4 = 0 or x^3 = 27
x = 0 or x = 3
To find the real-number solution of the equation 3x^7 = 81x^4, you are on the right track by setting the equation equal to 0 and factoring it. Let's continue from where you left off:
The equation can be rewritten as:
x^4 (3x^3 - 81) = 0
Now, we can further simplify this equation by factoring out the common factor x^4:
x^4 (3x^3 - 81) = 0
x^4 (3x^3 - 3^4) = 0
x^4 (3x^3 - 3^2 * 3^2) = 0
x^4 (3x^3 - 9 * 9) = 0
x^4 (3x^3 - 9^2) = 0
At this point, we notice that the expression inside the parentheses is a difference of squares:
x^4 (3x^3 - 9^2) = 0
x^4 (3x^3 - (3^2)^2) = 0
x^4 (3x^3 - 3^4) = 0
x^4 (3x^3 - 3^2 * 3^2) = 0
x^4 (3x^3 - 9 * 9) = 0
x^4 (3x^3 - 81) = 0
x^4 (3x - 9)(3x^2 + 3x + 9) = 0
Now we have factored the equation completely. To find the real-number solutions, we set each factor equal to 0 and solve for x separately.
First factor:
x^4 = 0
This implies that x = 0.
Second factor:
3x - 9 = 0
Solving for x: x = 3.
Third factor:
3x^2 + 3x + 9 = 0
This quadratic equation doesn't have real-number solutions because its discriminant (b^2 - 4ac) is negative.
Therefore, the real-number solutions of the equation 3x^7 = 81x^4 are x = 0 and x = 3.