Hi,
So I have the exponential expression (16x^6)^(1/2). I got the solution 4x^2, which I thought was correct. However, according to my textbook, the solution is 4|x|^2. Do anyone know why this is so?
I hope someone can help!
Hi,
So I have the exponential expression (16x^6)^(1/2). I got the solution 4x^2, which I thought was correct. However, according to my textbook, the solution is 4|x|^2. Do anyone know why this is so?
I hope someone can help!
Both you and your book seem to be incorrect.
$\left(16x^6\right)^{1/2} = 16^{1/2}(x^6)^{1/2} = 4|x^3|$
Let's use an example:
Suppose $x=-1$. Then $\left(16(-1)^6\right)^{1/2} = \sqrt{16\cdot 1} = \sqrt{16} = 4$, but $4x^3 = 4(-1)^3 = -4$. Therefore, you need to take the absolute value.
Suppose you have $y^2 = 16x^6$. Take the square root of both sides. You get: $y = \pm \sqrt{16x^6}$. That plus or minus is because when you square a real number, you get a nonnegative number, but the original number may have been positive or negative. The square root function always gives you the positive root, not the negative one. In other words, $\left(16x^6\right)^{1/2}$ is only giving you the positive root, so you take the absolute value.
But when I simplify, I only end up getting the square root of 16, not of x^6. As follows:
= (16x^6)^(1/2)
= 16^(1/2)x^6^(1/2)
= +/-√16 x^(6/2)
= +/-4 x^3
It only makes sense to me to include the positive and negative value of 4, since that came from 16 which I got the square root of. At what point do I take the square root of x^6 ?
When you multiply two numbers together, you get one number. If you multiply together 4 with $x^3$, you get one number. You know that the product must be positive (because the square root function only gives you the positive root). You know that $4$ is positive. The only other factor to consider is $x^3$. You are not taking the square root of $x^6$ at a separate time. It all happens at once. But, the end result must be positive. So, take the absolute value. If you want, you can put the whole thing inside an absolute value sign. The result is the same, you need to have a positive product. And the rules for multiplication inside of absolute value signs:
$|a\cdot b| = |a||b|$
So, applying that to the product:
$|4x^3| = |4||x^3| = 4|x^3| = 4|x|^3$
I think the idea of what you're saying is clicking. Please consider my thought-process behind this all:
We know that any number to an even exponent is positive, regardless of whether the number was negative or positive beforehand. What I'm saying is that before the expression (16x^6)^(1/2) is simplified, the variable of x becomes positive because it has an even exponent of 6 attached to it. For example,
= sqrt(x^6)
let's say x = 5^6 then,
= sqrt(5^6)
= sqrt(15625)
= 125
125 is 5^3... similar to how we are left with x^3 in the simplified expression of (16x^6)^(1/2). I know that 125 is positive because it got multiplied by an even exponent of 6. Therefore with this logic, I would believe that any variable you plug into 4x^3 in the simplified expression of (16x^6)^(1/2) would be positive. Unfortunately, this isn't the case because the simplified expression doesn't know any better to take into account the fact that it had an even exponent attached to it way back when it was in its original form ...which would of made it positive. This is essentially why you need to take the absolute value of the simplified expression, since we want for it to take into account the variable's true value, which is positive.
Is my thinking about this right?
Your expression and the textbook's solution are identical. However, both are wrong!
Inside the first set of parentheses you have 16 times X to the 6th power. 16 is a coefficient of X. 16 is not raised to the 6th power! Inside the second set of parentheses you have 1/2. The carat in front of the second set of parenthese is an alternate way of writing take what is inside the first set of parentheses and raise it to the 1/2 power. In other words, take the square root of what is inside the first set of parenthese.
To take the square root of what is inside the first set of parenthese, you have to take the square root of each factor inside the the first set of parenthese. This means taking the square root of 16 AND taking the square root of X to the 6th power separately. The square root of 16 is 4 and the square root of X to the 6th power is X to the 3rd power.
Your original expression simplified is therefore 4 x X to the third power not 4 times x to the second power as in the textbook!
Steve
Perhaps my effort here might reduce some of the the obfuscation that (to me) now appears to have crept in.
First, I need to say this . . .
You have: √(16x^6) . . . of which you have state that you have "found the solution".
But there is no solution to be found! There are only 'solutions' to equations. What you have written is not an equation: there is no 'equals sign' in it. What you do have is an arithmetical expression, which, presumably, you are required to simplify.
With that bit (an important 'bit', too) now out of the way, here is my reasoning to simplify that expression.
√(16x^6) can be written as √16 . √(x^6), where the '.' signifies multiplication.
√16 = ±4; √(x^6) can be written as √(x².x².x²), because (x^a).(x^b).(x^c) = x^(a+b+c).
So, √(x^6) = √(x².x².x²) = x√(x².x²) = x.x² = x^3.
Thus, √(16x^6) = ±4x^3.
Hope that helps.
Al.
@Skywave, I guess it falls to someone to keep things correct; it's me.
I have no idea what that sentence even says.
It is a gross mistake and mathematically ignorant statement to assert that ${\Large{\sqrt{16}=\pm 4}}$
The number $\Large{\sqrt{16}}$ is one number not two.
The expression $\pm 4$ indicates two numbers, one negative, one positive.
Back to the problem at hand. Consider $\large{f(x)=\sqrt{16x^6}~\&~g(x)=4x^3}$
The function $f$ has a domain of all real numbers and the range is set of non-negative real numbers.
The function $g$ has a domain of all real numbers and the range is set all real numbers.
Therefore $f\ne g,~~\sqrt{16x^6}\ne 4x^3$