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The first thing we need to understand is: what is a radical equation?. A radical equation is made up of a variable under a radical sign, numbers, possibly the same variable outside of a radical sign and an equal sign. A radical sign could be a square root, a cube root, a fourth root, etc. The small number 3√ to the left of the radical sign is called the index and will determine what it is...if it is a 3 like in this example it is a cube root. The opposite of a 3√ is to raise it to that power...in this case to the third power (or we say cube it). If it is a 4√ , then we raise it to the fourth power, etc. A √ without a number is a square root and its opposite is to raise it to the second power or square it.
So lets go over the steps used to solve a radical equation.
1) First get the radical sign by itself on one side of the =.
2) Then we do the opposite of the index on the radical sign to both sides of the equation.
3) Then we finish solving the equation by isolating the variable...or if there is an exponent on the variable, get 0 on one side and factor (see solving quadratic equations).
4) If there is more than one radical...isolate one, do the opposite... then isolate the other radical sign and do the opposite again and continue with solving.
5) Check your answer back into the original equation to make sure that it works. If it doesn't work, throw that answer out. If no answer is left, the answer is the empty set or no solution.
Lets start with a simple radical equation like: √x = 3. Since the radical is by itself, we are ready to raise it to the opposite power which is raising it to the second power or squaring it like this....(√x)2 = 32. What results is x = 9 because the square cancels the √ out and leaves us with only what is under the square root symbol. When we check it back in to the original equaion....√9 = 3...it works, therefore 9 is the correct answer.
Now lets try solving this radical equation :
| √(3x- 2) =7 |
Since the radical is by itself...we are ready to square both sides. |
| (√(3x- 2))2 = 72 |
Square both sides. |
| 3x-2 = 49 |
Squaring both sides cancels out the square root sign on the left and squares the number on the right. |
| 3x = 51 |
Then we finish solving the equation by isolating the variable. i.e. get x by itself. |
| x = 17 |
Divide by 3 |
| √(3(17)- 2) |
Then check your answer by substituting it back into the original equation to see if it works. |
| √(51-2) = 7 |
simplify
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| √49 = 7 |
It checks out...so 17 is the correct answer. |
Lets try a cube root radical equation.
| 3√(2x + 8) = 6 |
Again we can see that the 3√ is by itself. |
| (3√(2x + 8))3 = 63 |
So we cube both sides to cancel out the 3√ (or we raise it to the third power) |
2x + 8 = 216
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The 3√ cancels out on the left and the number gets cubed on the right. |
| 2x = 208 |
Then we finish solving for x again |
| x = 104 |
Divide |
| 3√(2(104) + 8) = 6 |
Check it by substituting 104 back in the original equation.
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| 3√ 216 =6 |
Again it works....so 104 is the correct answer. |
I think we are ready for a more difficult radical equation.
| 5 +√(x + 7) = x |
Notice this time the radical sign is not by itself, so we must first subtract 5 from both sides. This leaves us with: |
| √(x + 7) = x - 5 |
We subtracted 5 from both sides. |
| (√(x + 7))2 = (x - 5) 2 |
Now we square both sides. |
| x + 7=x2 - 10x + 25 |
Notice that squaring both sides cancelled out the square root on the left and "foil" the right (see simplifying polynomials) |
x2 - 11x +18 = 0
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This time instead of getting x by itself, we must get 0 on one side so we can factor the quadratic equation (see factoring) |
| (x - 9)(x - 2) = 0 |
Factored |
| x - 9=0 or x - 2=0 |
Set each parantheses equal to 0. |
| x = 9 or x = 2 |
Solve each smaller equation |
| 5 +√((9) + 7) = 9 |
First check 9....then check 2 back in the original equation. |
| 5 + √16 = 9 |
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| 5 +4 = 9 |
This checks so 9 one answer. |
| 5 + √((2) + 7) = 9 |
Now check 2 |
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5 + √9 = 2
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| 5 + 3 ≠ 2 |
2 does not work...therefore 2 isn't one of the answers. Therefore the only correct answer is 9. |
So now lets look at an equation that must be squared more than once.
| √x + √(x + 5) = 5 |
First we must isolate one of the square roots. |
| √x = 5 - √(x + 5) |
Subtract √(x + 5) from both sides. |
| (√x)2=(5 -√(x +5))2 |
Square both sides. |
| x = 25 -2*5(√(x + 5)) + (x + 5) |
This cancels the square root on the left...but we must "foil" the right side.
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-30 = -10(√(x +5)) |
Add 25 and 5 together and subtract it from both sides. Subtract x from both sides and it cancels out on both sides. |
3 = √(x + 5)
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Divide both sides by -10. |
| 32 = (√(x + 5))2 |
Square both sides. |
9 = x + 5
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Subtract 5 from both sides. |
| √4 + √(4 + 5) = 5 |
Check 4 back into the original equation to see if it makes the equation true. |
| 2 + 3 = 5 |
Since it works in the check, 4 is the correct answer. |
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