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Pre-Algebra - Simple Adding and Subtracting Radicals
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In order to add radicals, there must be “like” terms in the problem. A problem with like terms would be 5x + 10x. Both terms contain the variable “x”. So the solution to 5x + 10x = 15x.
Let’s look at the next square root problem of 6√2 + 8√2. The like term here is √2. In addition to the √2 being like terms, so are the coefficient numbers “6” and “8” so you get to add those together, i.e., 6 + 8 = 14. The 14 is then placed before √2 and the solution is 14√2.
The same rule holds true for subtraction.
There must be “like” terms in the problem. Knowing this, if we look at the problem of: 9√p - 3√p the like term is √p and the coefficient numbers are 9 - 3 = 6. The 6 is then placed before the like term of √p giving us the solution of 6√p.
Now let’s look at another problem. 10√12 + 2√3. You cannot add these binomials because the terms are not like terms. However, you can simplify 10√12 to read 10√4 ● 3 and then the number 4 is the perfect square root of 2 so the 2 moves to the left of the √ symbol so that the problem now looks like: 10 ● 2√3. Next, 10 ● 2 = 20 giving you 20√3. Now we go back to the original problem and rewrite is as 20√3 + 2√3. We have √3 as our like term and then 20 + 2 = 22 which is placed before our like term giving us the solution of 22√3.
For each radical expression shown in the questions below, add or subtract to find the solution.
Solution: 19√9
Answer (b) is the correct solution
Solution: 14√3
Answer (d) is the correct solution
Solution: 59√x
Answer (a) is the correct solution
Solution: 35√12
Answer (b) is the correct solution
As we do not have like terms we must simplify as follows:
6√125 = 6√25 ● 5
As 25 is a perfect square in that 5 ● 5 = 25, the 5 goes before or to the left of the √ symbol giving you 6 ● 5√5
6 ● 5 = 30 and then add on the √5 giving you 30√5. Now the problem can be rewritten as: 30√5 - 4√5 = 26√5
Solution: 26√5
Answer (c) is the correct solution
As we do not have like terms we must simplify as follows:
10√48 = 10√16 ● 3
As 16 is a perfect square in that 4 ● 4 = 16, the 4 goes before or to the left of the √ symbol giving you 10 ● 4√3
10 ● 4 = 40 and then add on the √3 giving you 40√3. Now the problem can be rewritten as: 40√3 + 15√3 = 55√3
Solution: 55√3
Answer (b) is the correct solution
As we do not have like terms we must simplify as follows:
√144 has a square root as 12 ● 12 = 144. Therefore, the square 12 is then placed before or to the left of the √ symbol giving us 12√
Next, √64 also has a square root as 8 ● 8 = 64. The square 8 is then placed before or to the left of the √ symbol giving us 8√
We now have like terms of √ and the problem can be rewritten as:
12√ - 8√ = 4√
Solution: 4√
Answer (c) is the correct solution
Solution: 12a√pq
Answer (d) is the correct solution
Solution: 100√xy
Answer (a) is the correct solution
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As we do not have like terms we must simplify as follows:
45√72 = 45√36 ● 2
As 36 is a perfect square in that 6 ● 6 = 36, the 6 goes before or to the left of the √ symbol giving you 6 ● 45√2
6 ● 45 = 270 and then add on the √2 giving you 270√2. Now the problem can be rewritten as: 270√2 - 221√2 = 49√2
Solution: 49√2
Answer (c) is the correct solution