SOLUTION: Solve for x.
x^(3/2) - 3x^(1/2) = 0
sqrt{x^2} - sqrt{3x} = 0
After squaring both sides, I got x^3 = 3x.
x^3 - 3x = 0
x(x^2 - 3) = 0
x = 0 and x^2 - 3 = 0
Solvi
Algebra.Com
Question 1199216: Solve for x.
x^(3/2) - 3x^(1/2) = 0
sqrt{x^2} - sqrt{3x} = 0
After squaring both sides, I got x^3 = 3x.
x^3 - 3x = 0
x(x^2 - 3) = 0
x = 0 and x^2 - 3 = 0
Solving x^2 - 3 = 0 for x, I got sqrt|3}, -sqrt{3}.
After checking my answers for x, I concludes that the only two solutions
for x are 0 and sqrt{3}.
The textbook answers for x are 0 and 3.
Why am I wrong?
Found 3 solutions by ikleyn, greenestamps, math_tutor2020:
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.
You are wrong starting from the third line in your post.
--------------
Comment from student : Why is my third line wrong?
My response : If you ask such questions, it means that you do not know the necessary basics.
In other words, you do not have a necessary prerequisites to solve such problems.
In short, you are in wrong class.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
On your second line you have replaced with .
Those two expressions are not the same.
Note if you solve by squaring both sides of the equation, you will need to check for extraneous solutions. Solve by factoring instead.
or
ANSWERS: 0 and 3
------------------------------------------------------------------
added in response to reader's question....
Extraneous solutions are solutions to an equation you end up with in your work that don't satisfy the original equation.
The most common place where this can happen is if, in the process of solving the original equation, you square both sides at some point. The extraneous solutions are the result of the fact that the square of a negative number is positive.
For a very simple example, consider the equation
We know immediately that there is no solution, because a square root is never negative. But when we square both sides...
But x=9 does not satisfy the original equation. It is an extraneous root, introduced when we squared both sides of the original equation.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
The is not the same as
In the first expression, we apply the 1/2 exponent to only the x; while the second expression has the exponent apply to all of 3x.
Recall that with PEMDAS, the parenthesis is done first, then exponents, then multiplication.
P = parenthesis
E = exponents
M = multiplication
D = division
A = addition
S = subtraction
The reason I bring this up is that with , we compute the 3x first in the parenthesis, then apply the exponent later.
Whereas with , we do the exponent piece first before multiplying with the 3 out front.
-----------------------------
Here is one way to solve for x.
Rearrange exponents in the first term. Use the rule a^(b*c) = (a^b)^c.
Raising to the 1/2 power is the same as a square root.
Square both sides
or or
or or
Those are the potential solutions.
But we need to check each of them.
-----------------------------
If you were to plug in x = 0 into the original equation, then,
We end up with a true statement, therefore confirming x = 0 is indeed a root.
-----------------------------
Repeat for x = 3
We have confirmed x = 3 is also a root.
-----------------------------
Now try x = -3
We see that a negative is under the square root, to produce a complex value in the form a+bi, where
So we'll ignore x = -3.
-----------------------------
Another approach to take is to use a factoring method similar to what @greenestamps mentioned.
or
or
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