How to solve equations with square roots, cube roots, etc.
Radical Equations
A Radical Equation is an equation with a square root or cube root, etc. |
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Solving Radical Equations
We can get rid of a square root by squaring. (Or cube roots by cubing, etc)
But Warning: this can sometimes create 'solutions' which don't actually work when we put them into the original equation. So we need to Check!
Follow these steps: Elements lab 3 3 2.
- isolate the square root on one side of the equation
- square both sides of the equation
Then continue with our solution!
Example: solve √(2x+9) − 5 = 0
square both sides:2x+9 = 25
Now it should be easier to solve!
Divide by 2:x = 16/2 = 8
Check: √(2·8+9) − 5 = √(25) − 5 = 5 − 5 = 0
That one worked perfectly.
More Than One Square Root
What if there are two or more square roots? Easy! Just repeat the process for each one.
It will take longer (lots more steps) .. but nothing too hard.
Example: solve √(2x−5) − √(x−1) = 1
isolate one of the square roots:√(2x−5) = 1 + √(x−1)
We have removed one square root.
expand right hand side:2x−5 = 1 + 2√(x−1) + (x−1)
subtract x from both sides:x−5 = 2√(x−1)
Now do the 'square root' thing again:
square both sides:x−1 = ((x−5)/2)2
We have now successfully removed both square roots.
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Let us continue on with the solution.
It is a Quadratic Equation! So let us put it in standard form.
Multiply by 4 to remove division:4x−4 = x2 − 10x + 25
Combine like terms:−x2 + 14x − 29 = 0
Using the Quadratic Formula (a=1, b=−14, c=29) gives the solutions:
2.53 and 11.47 (to 2 decimal places)
Let us check the solutions:
2.53: √(2×2.53−5) − √(2.53−1) ≈ −1 Oops! Should be plus 1.
11.47: √(2×11.47−5) − √(11.47−1) ≈ 1 Yes that one works.
There is really only one solution:
Answer: 11.47 (to 2 decimal places)
See? This method can sometimes produce solutions that don't really work!
The root that seemed to work, but wasn't right when we checked it, is called an 'Extraneous Root'
So: Checking is important.
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