Question 695221: Directions say to solve the Equation . But I have Trouble with These, their Confusing.
3 radical x = radical 6x+4
Found 2 solutions by stanbon, RedemptiveMath:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 3 radical x = radical 6x+4
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3sqrt(x) = sqrt(6x+4)
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Square both sides:
9x = 6x+4
3x = 4
x = 4/3
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Checking the answer:
3sqrt(4/3) = sqrt[6(4/3) + 4]
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3sqrt(4/3) = sqrt[8+4]
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3sqrt(4/3) = 2sqrt(3)
3sqrt(12)/3 = 2sqrt(3
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sqrt(12) = 2sqrt(3)
2sqrt(3) = 2sqrt(3)
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x = 4/3 is a solution.
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Cheers,
Stan H.
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Answer by RedemptiveMath(80) (Show Source):
You can put this solution on YOUR website! One of the best methods to solving variables in radicals is to get rid of the radicals first. This will give us a clearer understanding of what we are dealing with. So, given our problem 3√x = √(6x+4) (I'm assuming 6x+4 is all underneath the second radical), we need to get rid of the radicals first. But how would we do that?
So far in algebra you've learned how to solve for variables in equations without radicals. For example, if you had the equation 9x + 4 = 40, you'd know that x = 4 because you subtracted 4 from each side and divided each side by 9. This would give you an answer of x = 4. What we are really doing when solving for variables in equations is working the problem backwards. That is, we are using operations that are opposites of what is going on in the given equation. We subtracted 4 in the equation 9x + 4 = 40, but the operation in the equation isn't subtraction (4 is being added in the equation, not subtracted). Then we divided by 9 to get x by itself, but the equation reads "9 times x." We can see that we are doing opposite of what the given equation states. We are undoing the equation. The same goes for radicals. Let's us analyze another problem.
Let's say we had any equation 3x^2 = 243. You would know from algebra that we would need to divide 3 from each side first to get x^2 = 81. Then we would need to square root both sides to get x without an exponent. We receive the answer x = 9 (or -9). Notice how this problem relates with our present one. This problem gives exponents first (in the equation 3x^2 = 243), but we eventually use a square root operation to get x by itself. In our problem, to get rid of the radicals we would need to use an exponent operation. That is, we would need to square both sides since they are being square rooted first. So,
3√x = √(6x+4)
(3√x)^2 = [√(6x+4)]^2.
Notice how everything in both sides is being squared, not just the things under the radical. We must remember that if we do something to a part of the equation, we must do it to all of the equation. Let us further examine our problem:
(3√x)^2 = [√(6x+4)]^2
(3)^2(√x)^2 = [√(6x+4)]^2.
Since 3 is outside of the radical in the first part of the equation, it must be squared by itself. If you remember law of exponents, if we have (9x)^2, it would become (9^2)(x^2) or (9x)(9x). The square operation means whatever is being squared is being multiplied by itself (the terms are being multiplied to each other twice). Solving for this expression, we would get 81x^2. We could write our problem as (3√x)(3√x) instead of (3)^2(√x)^2, but we would need to know how to do the first operation. We multiply everything outside of the radicals together, and we multiply everything inside the radicals together when we multiply. So,
(3)^2(√x)^2 = 9(√(x^2)) = 9x.
And if you remember the properties of exponents and radicals, then you can see that if something is being squared underneath a square root operation, the exponent and the radical cancel each other out. So, √(x^2) simply becomes x. We could look at this using numbers and see why we can just cancel the square root and the exponent out:
(√9)^2 = (√9)(√9) = √81 = 9.
NOTE: If you've talked about indexes, then we must keep them in mind. This cancelling out will only work if the exponent and the index of the radical are the same. The square root symbol has an implied index of 2, and when the squared operation is being applied to subjects underneath (respective circumstances), then the two can cancel out. Cubed roots are cancelled out with cubed operations, and so forth.
So, we are left with 9x on the left side of the equation. Now we need to finish the square operation on the right side:
[√(6x+4)]^2
[√(6x+4)][√(6x+4)]
6x + 4.
Using the rules above, we can just cancel the square root symbol and squared symbol out. Now we are left with a clearer picture of the equation:
9x = 6x + 4
3x = 4
x = 4/3.
To check ourselves, we must plug x = 4/3 into our original equation:
3√x = √(6x+4)
3√(4/3) = √[6(4/3)+4]
3√(4/3) = √(8+4)
3√(4/3) = √12
3(1.1547) = 3.4641
3.4641 = 3.4641.
I used a calculator to find the approximation of each square root. We could leave the radicals in this work and just simplified. We would have still found that the each side equaled the other.
Below is a list of PEMDAS (order of operations) and its opposites we use all the time when we solve for equations. The steps to using them in equations (for PEMDAS) or using to them undo equations (solving for the variables) are listed.
PEMDAS:
1. Parentheses (everything in them)
2. Exponents
3. Multiplication/Division (generally from left to right)
4. Addition/Subtraction (generally from left to right)
Solving for variables:
1. Addition/Subtraction (loose numbers not side by side of variables)
- If addition in equation, use subtraction.
- If subtraction in the equation, use addition.
2. Multiplication/Division (terms on variables)
- If multiplication in the equation, use division.
- If division in the equation, use multiplication.
3. Exponents (on variables)
- If exponents are used in the equation, use their respective root operations.
- If root operations are used, use their respective exponent operations.
As you can see, the steps for writing/solving an equation are practically opposite. However, you must be careful with parentheses when undoing an equation. You have to analyze an equation in order to know what do to with parenthetical quantities. It may turn out that you don't have to worry about the parentheses when undoing an equation (in some cases).
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