Question 1164447: The problem is 8x^3=125. Why can't I simply take the cubed root of both sides to get 2x=5, and from there get x=5/2?
The solutions I have seen require I move all terms to the left and set the equation equal to zero, factor using difference of cubes formula, and solve from there, giving me x= 5/2, and x=-5+/-5i(square root of 3) all over 4. I don't understand why I can't just solve it as the equation is given, instead of setting everything to 0. Thank you!
Found 3 solutions by ikleyn, MathTherapy, greenestamps:
Answer by ikleyn(52814)   (Show Source):
You can put this solution on YOUR website! .
In this concrete case, you can.
In this case, until you restrict yourself by real numbers, you can.
In more complicated cases, the cubic root from a number is a complex number, and even not one complex number,
but THREE, instead.
So, in this complex case, after taking the cubic root, you may have 3 values of the cubic root on the left side
and 3 cubic root values on the right side --- and how then you will decide who is equal to whom ?
It is the story . . .
/\/\/\/\/\/\/\/
After reading the post by tutor @MathTherapy.
When you take the CUBE ROOT, you DO NOT NEED to write +-.
Until you work with real numbers, keep the sign of the resulting real number (cube root)
SAME as the number UNDER the cube root sign.
The post by @MathTherapy is a mistake.
Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website! The problem is 8x^3=125. Why can't I simply take the cubed root of both sides to get 2x=5, and from there get x=5/2?
The solutions I have seen require I move all terms to the left and set the equation equal to zero, factor using difference of cubes formula, and solve from there, giving me x= 5/2, and x=-5+/-5i(square root of 3) all over 4. I don't understand why I can't just solve it as the equation is given, instead of setting everything to 0. Thank you!
No, you CANNOT! SORRY, but you won't be able to obtain all answers if you do that!
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Let's look at a simpler, more familiar example, to see what the problem is with just taking the cube root on both sides.
Suppose the equation is
If you just take the square root of both sides, you get x=2. But you know there are two solutions, 2 and -2.
Algebraically, to get all the solutions (in this case, both), you need to set everything equal to 0 and solve.
x=2 OR x=-2.
In that case, both solutions are real numbers. In cases of equations with higher powers of x, that is often/usually not the case. A polynomial equation of degree n has n roots; only in rare special cases are all the roots real.
In your example, of the form x^3=A where A is any real number, there is always only one real root.
So when you solve the equation by taking the cube root of both sides, you find only the real root; and you have lost the information you need to find the non-real roots.
To find all three roots, you need to use the pattern for factoring the difference of cubes. That will give you a linear factor which gives a real solution and a quadratic factor that gives two non-real solutions.
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