The
Purplemath Forums 
Solving Radical Equations: Examples (page 5 of 6)
Square both sides, being careful to write out the square on the righthand side: Then x = –8 and x = 2. Are both of these solutions valid? Graphing the lines for either side of the original equation: ...I get the following graph: It appears that both solutions are valid. Here's the check:
So the solution is x = –8 or x = 2. The following examples are not complete. I'll leave the checking to you!
This equation will have to be squared twice in order to solve it: It appears that the solutions are x = –5 and x = 0. However, only one of these solutions is actually valid. To find out which one, check them both.
This equation will also have to be squared twice. Don't forget to square that 3 in front of the square root on the righthand side! Copyright © Elizabeth Stapel 20022011 All Rights Reserved To solve this, use the Quadratic Formula. Then check your answers, because only one is actually valid.
This one is actually simpler than the two previous examples, because the two square roots are multiplied together, rather than added or subtracted. So this equation will need to be squared only once: Then the solutions are
x = –9
and x
= 16. But x
cannot equal –9,
because this would put negatives inside both radicals in the original
equation. Now you check the other solution, to see if it might work.
Since there is a square root inside a square root, I'll have to square twice: Using the Quadratic Formula, I get solutions of x = ^{401}/_{144} and x = 3. Check these, as only one is a valid solution. << Previous Top  1  2  3  4  5  6  Return to Index Next >>



Copyright © 20022012 Elizabeth Stapel  About  Terms of Use 




