WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In previous lessons, we have learned techniques for working with radicals.
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Now, we are going to put those together and use them to solve radical equations and inequalities.
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First, the definition of a **radical equation**: recall that radical equations contain radicals with variables in the radicand.
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For example, it is something like the fourth root of x + 5 equals 3, or 6 + √(2x + 1) = 4.
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These are both examples of radical equations.
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If you had a radical, and there was just a number under it, like the square root of 5, plus 2x,
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this is not a radical equation, because there is no variable under the radical sign.
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OK, in order to solve these equations, we are going to use a technique learned earlier on in Algebra I and reviewed now, and also go into more complex problems.
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For example, if we had the square root of x + 2, minus 6, equals -1; what we are going to do is
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raise each side of the equation to a power that is the index of the radical.
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So, the index of the radical here is 2; so I am going to raise both sides to the second power; in other words, I am going to square both sides.
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But before you do that, first isolate the radical; then raise both sides to a power equal to the index of the radical.
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First, I am isolating this by adding 6 to both sides; so, -1 + 6 is going to give me 5.
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Now, I am going to square both sides; I am going to raise both sides to that index power, which is 2.
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Well, since the index is 2, and it is being raised to the second power, this equals the radicand, and this equals 25.
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Subtracting 2 from both sides gives me x = 23.
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So, to solve radical equations, you isolate the radical, then raise both sides to a power equal to the value of the index; then, solve the equation.
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Now, you see here that it says sometimes this must be done twice in order to eliminate all radicals.
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That is a more complex situation that we are going to look at right now.
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Let's say I am given √(x + 3) - 2 = √(x - 5).
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I can't just isolate the radical, because there is more than one radical.
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So in this case, I just begin by raising both sides to a power that is equal to the index.
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Since the index is 2, I am just going to first square both sides; I can't isolate.
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Squaring both sides is going to give me this; and if I look at this left side, I am squaring a difference; that is analogous to this situation:
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(a - b)², which turns out to be a² - 2ab + b².
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I just substitute in here; a is the square root of x + 3, and b is 2.
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So, that makes my work on the left much quicker, instead of having to use FOIL, use the distributive property,
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multiply everything out, add...I just say, "OK, the square root of x + 3, squared, minus 2 times a times b..."
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a is the square root of x + 3, and then b is 2.
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And then, finally, I get b².
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On the right, I raise this to the index power, so I get the radicand, x - 5.
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OK, I squared the square root of x + 3; the square root of (x + 3)² is simply the radicand, x + 3.
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Here, I have 2 times 2 is 4, times the square root of x + 3.
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Here I have -2 times -2 is 4.
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Now, you can see that the radical on the right was eliminated.
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However, I still have a radical on the left; so that is why I am going to need to repeat this process.
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I squared both sides, and then I need to repeat if radicals remain.
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I squared both sides, and I am going to need to repeat; let's simplify first.
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This is 3 + 4, so this gives me x + 7, minus 4√(x + 3), equals x - 5.
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I am going to move this 7 over to the right by subtracting 7 from both sides; and that is going to be x - 5 - 7, so x - 12.
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Now, the next thing is: to isolate this, I want to subtract x from each side; so x - x...that drops out.
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x -x...that drops out, too: conveniently, the x's dropped out, making this even simpler to work with.
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I still want to isolate this completely; so, to do that, I need to divide both sides by -4.
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Now, I am left with something easier to work with, because I end up with just the radical isolated on the left, and then I just have a number on the right.
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So, I am going to repeat my process, and I am going to square this; and let's start right up here with that second process, √(x + 3)² = 3².
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I raised this radical to the index power; that is going to leave me with the radicand, x + 3; and on the right, I have 9.
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This gives me x = 6; this was pretty complex.
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And I solved it by first squaring both sides, which got rid of the radical on the right.
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Then, I simplified the left, found that the x's dropped out, combined my constants, and ended up with this.
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To completely isolate the radical, I divided by 4; and when I ended up with this, then I have something that is more familiar,
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where I have an isolated radical expression on the left, and squared both sides, solved, and determined that x equals 6.
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OK, recall that sometimes, when we use this method of squaring both sides, we end up with a solution;
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and when we check the solution, it actually does not satisfy the original equation.
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This is a result of the squaring step of using this technique; and these solutions are called extraneous solutions.
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Therefore, you always must check the solutions you get from this technique back in the original equation or inequality.
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If the solution exists, and you use this technique, the solution will be among your results.
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And I say "among your results" because you might have the valid solution, plus some extraneous solutions.
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There may be no solution; you may just end up with one or two extraneous solutions; you may end up with one or more valid solutions.
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There is no way of knowing; but if the solution exists, it will be somewhere among the solutions you end up with.
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The only way to find it, though, is to check all the solutions to see what you have.
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For example, we did, earlier on, that example (which was a little bit complicated): √(x + 3) - 2 = √(x - 5).
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And when we worked that out, I came out with the solution x = 6.
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I can't just assume it is a valid solution; it actually may not be valid.
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The only way to know is to go back and substitute in.
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Everywhere there is an x, I am going to make it a 6; and I am going to see if my equation holds up.
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This is going to give me √(x + 3), which is √9, minus 2, equals √(6 - 5); so that becomes √1.
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The square root of 9 is 3; minus 2 equals the square root of 1, which is 1.
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1 does equal 1; this is true, so the solution, which is x = 6, is valid.
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It is possible sometimes that you will get a solution, and you check it, and you get something strange like 4 = 5, which is not true.
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That means that this answer, this solution, does not satisfy the equation; and therefore, it was an extraneous solution
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that popped up as a result of squaring both sides.
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I could have ended up with two solutions or three solutions; and some were extraneous/some were valid/none were valid.
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Again, the only way to tell is to check.
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Eliminating nth roots: we have been doing this, but just to bring it out and explain exactly what we were doing:
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to eliminate an nth root, you isolate the root (I talked about isolating the radicals),
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and then raise both sides of the equation to the power n.
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That is what we did when we took something like √(x + 2) - 4 = 8, and we first isolated, which would give me 12.
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And then, I said, "OK, n is 2, so I am going to square it; I am going to raise it to the second power."
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So, this is exactly what we have been doing: isolate the root, and then raise both sides of the equation to that index power.
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And the same technique could apply to if you are using a higher index; but it does become more complicated.
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OK, and again, an important last step is to check solutions in the original equation to make sure that they are valid.
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They may be valid; they may be extraneous.
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All right, we talked about radical equations; and radical inequalities are very similar.
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So, these are inequalities with variables in the radicand.
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And just as we talked about with equations (radical equations have variables in the radicand), radical inequalities
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are the same thing, only we are working with greater than or less than, or greater than or equal to, and all that, instead of equals.
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We have this restriction that we have seen before with radicals: if the index is even,
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we have to determine the values of the variable for which the radicand is greater than or equal to 0.
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In other words, we have to restrict the solution set to values that will not make the radicand negative,
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because if the radicand is negative, and we have an even index, we will end up with something that is not a real number.
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So, after we determined that restriction, we solved algebraically, using the same techniques that we have talked about before with radicals.
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Let's use an example: the square root of 3x + 4, minus 2, is greater than 5.
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Before I even begin trying to solve this, I am going to figure out what my excluded values are.
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So, I know that 3x + 4 has to be greater than or equal to 0.
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If this ended up being negative, and I have an even index here, 2, then I can end up with a complex number as a solution.
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And I don't want that; we are just working with real numbers.
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So, I am going to go ahead and say 3x, and I am going to subtract 4 from both sides to get 3x ≥ -4.
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And then, I am going to divide both sides by 3; so x ≥ -4/3.
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This is just saying that I am restricting the domain to these values.
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Any value of x that is less than -4/3 is an excluded value.
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So, x less than -4/3 are excluded.
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I haven't even begun to solve this yet; I was just figuring out what the restrictions are on my solution set.
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OK, so now, I am going to go ahead and actually solve this algebraically, using techniques that we used with radical equations.
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The first one here is to isolate the radical on one side of this inequality.
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And I am going to do that by adding 2 to both sides.
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So, this is going to give me √(3x + 4) > 7.
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Next, I am going to raise both sides to the power equal to the index, which, in this case, would be the power 2.
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So, squaring both sides is going to give me...the square root squared is going to give me the radicand,, 3x + 4 > 7 squared, or 49.
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Now, I just solve: 3x...subtracting 4 from both sides, the 4 drops out from here...49 - 4 is 45.
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3x > 45; and I want to isolate the x, so I am going to divide by 3; 3 is going to drop out--I get x > 45/3, or x > 15.
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And I check and say, "OK, is this allowed?"
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Yes, it is, because if x is greater than 15, it is going to be greater than -4/3.
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So, I am not going into values that are not allowed.
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And then, you could check this by picking a value greater than 15,
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and putting it in here, back into the original, and making sure that the inequality is true--that it holds up.
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I could pick a value such as 20 or 16, put it in here, solve this, and make sure that I ended up with something greater than 5,
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because again, when we use this technique of squaring both sides, we need to check the answers.
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OK, first example: I am going to solve this radical equation by the technique of first isolating the radical.
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So, all I have to do in order to achieve that is subtract 8 from both sides to get √(3x - 2) = 7.
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Then, I am going to square both sides, because the index is 2.
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So, if I raise both sides to the second power, then I am going to get the radicand from here,
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because the square root of 3x - 2, times itself, is 3x - 2.
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On the right, I get 7², is 49.
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Now, I just solve using my usual algebraic techniques, adding 2 to both sides.
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3x = 51; now, dividing both sides by 3, x = 51/3; and if you do the arithmetic on that, you will find that x equals 17.
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Another important step is checking to make sure this is a valid solution.
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So, I am going to go all the way back to this original; and I am going to let x equal 17
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and substitute that in to get 8 + √3 times 17 minus 2 equals 15.
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3 times 17 is 51; 51 minus 2 is 49; so, this is going to give me 8 + √49 (which is 7) = 15.
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Since 15 = 15, then x = 17 is valid, because x = 17 actually satisfied this equation--the two sides were equal.
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OK, the second example is a similar idea; but instead of working with square roots, now the index is 4.
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But I am going to use the same technique: I am going to first isolate this radical.
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And I am going to do that by subtracting 4 from both sides.
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6 - 4 is 2, so now I have that the fourth root of 2x - 8 equals 2.
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In this case, since the index power is 4, I am going to raise both sides to the fourth power.
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OK, I have the fourth root of 2x - 8 raised to the fourth power.
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If the index and the power you are raising it to are the same, this, then, equals the radicand.
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2 to the fourth power is...2 times 2 is 4, times 2 is 8, times 2 is 16.
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Next, I just isolate the x, using algebraic techniques: I am going to add 8 to both sides--that gives me 2x = 24.
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Dividing both sides by 2 gives me x = 12.
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And we can't forget the important step of checking our solution back in the original equation.
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I am going to check, and this is the fourth root of 2...I am going to check x = 12.
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The fourth root of 2 times 12, minus 8, plus 4, equals 6; let's see if that holds up.
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2 times 12 is 24, minus 8, plus 4, equals 6.
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This is going to give me the fourth root of 24 - 8, which is 16.
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The fourth root of 16 is 2, because 2⁴ = 16; so the fourth root of 16 is 2.
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That leaves me with 2 + 4 = 6; and 6 does equal 6, so x = 12 is valid.
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It is a valid solution, because it satisfies this equation.
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When x is 12, the equation holds true.
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In our third example, we have an inequality: we have to take that extra step of excluding values--finding what the excluded values are.
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3x - 6 must be greater than or equal to 0; it needs to be a non-negative number,
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because since this is an even power, if this radical is the square root of a negative number,
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then I could end up with an imaginary number, which I don't want.
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So, since this is an even index, I am going to take the extra step and define the values of x that I am allowed to work with.
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OK, 3x - 6 must be greater than or equal to 0; adding 6 to both sides gives me 3x ≥ 6.
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Now, dividing both sides by 3, x ≥ 2.
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So, I have not solved it yet; all I have said is that the only values that are even allowable to use are these.
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Now, I am going to go about solving it; so let's rewrite this here.
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2 plus the square root of 3x minus 6 is less than or equal to 8.
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Just like with radical equations, I start out by isolating the radical.
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So, I am going to subtract 2 from both sides; my index value is 2--it is just the square root.
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So, I am going to square both sides.
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The square root of something squared is just that value; so this becomes the radicand, because the square root of 3x - 6, times itself, is 3x - 6.
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It is less than or equal to 36.
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Now, I am going to add 6 to both sides: 36 + 6 is 42.
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Divide both sides by 3: 42 divided by 3 is actually 14, so x is less than or equal to 14.
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I am not quite done: I need to put all of this together, because if I just said,
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"OK, x is less than or equal to 14," I might say, "Oh, OK, x can be 0," but it actually can't,
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because I have this restriction right here on the domain that it has to be greater than or equal to 2.
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Therefore, I am going to write out the whole thing together, saying that x needs to be greater than or equal to 2
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in order to even be a value that will give me a real number.
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But to actually solve this inequality, to find the solution set...the solution set only encompasses those values of x that are less than or equal to 14.
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So, in this case, I actually had to put this all together.
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Now, in order to check this, one way to check it is to choose a value that is within the solution set (for example, 5--
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5 is greater than or equal to 2, and less than or equal to 14).
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So, I am going to check with x = 5, and I am going to see if that satisfies this inequality.
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2 + √(3x - 6) ≤ 8: if I am going to let x equal 5, and just make sure that this thing holds up,
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this is going to give me 15 - 6 under the radical, which is...15 - 6 is 9, so I have the square root of 9 there,
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which is 3; so I end up with 5 ≤ 8, and this is true.
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So, this helped to verify that I actually have a valid solution.
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Again, when you are using this technique of squaring both sides, extraneous solutions can occur.
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OK, this time, I actually have two radicals; so it is not going to be possible to just isolate the radical.
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Therefore, my first step is just going to be to get rid of at least one of the radicals by squaring both sides.
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So, I am going to go ahead and square both sides, and recall that, on the left, if I have something
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such (a - b)², it is going to give me a² - 2ab + b².
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And by remembering that, I will save myself the work of having to use the distributive property and figure this whole thing out.
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I am just going to go ahead and say, "OK, a equals the square root of x plus 21, and b equals 2."
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So, on the left, I am going to get a²...the square root of x plus 21, squared...minus 2 times a
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(the square root of x plus 21), times b (which is 2); and then, for the last term, I will get b², which is 2².
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On the right, I have a -1 in front of this; when I square that, -1 times -1 becomes 1.
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So, I am going to end up with a positive expression on the right, and the square root of x + 5 squared...
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since the index is the same as the power, this becomes the radicand; so on the right, I end up with x + 5.
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OK, simplifying: over here, I have a square root raised to the second power; the square root squared gives me the radicand.
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So, I can see I have gotten rid of this radical on the right; and this is not a radical.
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However, in the middle, this middle term is -2, times 2 is -4, times the square root of x plus 21.
00:26:40.000 --> 00:26:52.400
I am still left with this radical; here, 2 times 2 is 4; simplify before you proceed.
00:26:52.400 --> 00:27:02.000
This gives me 2 and 4, so that is x + 6 minus this equals x + 5.
00:27:02.000 --> 00:27:22.600
So, I am going to go ahead and subtract this 6 from both sides...this is actually 21.
00:27:22.600 --> 00:27:29.500
So, go ahead and simplify this; this is 21 plus 4, is actually going to be 25.
00:27:29.500 --> 00:27:42.000
So, x + 21, the square root of that squared, gives me x + 21; so that is 25 right here; 21 + 4 is 25.
00:27:42.000 --> 00:27:55.300
OK, simplify by subtracting 25 from both sides: 5 - 25 is -20.
00:27:55.300 --> 00:28:07.300
In the next step, I would subtract x from each side; so when I do that, I get x - x; the x drops out of the left.
00:28:07.300 --> 00:28:13.300
When I take x - x, the x also drops out of the right side.
00:28:13.300 --> 00:28:18.600
Now, I have gotten rid of this radical on the right; on the left, I have a radical,
00:28:18.600 --> 00:28:26.600
but I also have a -4 in front of it, so I need to isolate the radical by dividing both sides by -4.
00:28:26.600 --> 00:28:37.600
So, this gives me -20 divided by -4 on the right; so the square root of x + 21 equals 5.
00:28:37.600 --> 00:28:41.200
Since I still have a radical left, I have to go through this process again.
00:28:41.200 --> 00:28:54.400
So, let's take this up here and repeat the process of squaring both sides.
00:28:54.400 --> 00:29:03.300
The square root of x + 21, squared, gives me the radicand: x + 21 = 5²...that is 25.
00:29:03.300 --> 00:29:15.200
Subtracting 21 from both sides gives me 4; so I came up with this solution that x = 4, and I need to check that in the original.
00:29:15.200 --> 00:29:22.800
So, check by inserting 4 for each x in the original.
00:29:22.800 --> 00:29:35.800
That is going to give me the square root of 4 + 21, minus 2, equals negative...and then that is the square root of 4 + 5.
00:29:35.800 --> 00:29:43.500
This gives me the square root of 25, minus 2, equals minus the square root of 9.
00:29:43.500 --> 00:29:55.400
The square root of 25 is 5, minus 2 equals minus the square root of 9, which is 3; so that gives me -3.
00:29:55.400 --> 00:30:06.900
This is 5 - 2 = -3; this is not true; therefore, the solution is not valid.
00:30:06.900 --> 00:30:08.900
This solution, x = 4, is not valid.
00:30:08.900 --> 00:30:15.200
Now, I said that, when you use this method, if there is a valid solution, you will come up with it.
00:30:15.200 --> 00:30:19.800
You might have some extra solutions, but you will have a valid solution, if it exists.
00:30:19.800 --> 00:30:31.700
What this tells me, since this is not valid, is that there is no valid solution here.
00:30:31.700 --> 00:30:39.900
So, this was a pretty complex problem: there were two radicals, so we had to square to get rid of the radical on the right;
00:30:39.900 --> 00:30:47.100
do a bunch of simplifying; isolate the remaining radical, which is on the left;
00:30:47.100 --> 00:30:52.900
and repeat the process with that by squaring both sides to come up with x = 4.
00:30:52.900 --> 00:31:00.300
And then, after all that work, we went and checked it, and found that x = 4 does not satisfy this equation--
00:31:00.300 --> 00:31:04.800
that, when we use that, we end up with 3 = -3.
00:31:04.800 --> 00:31:10.900
Had that negative not been there, we could have come up with a valid solution.
00:31:10.900 --> 00:31:14.800
But with that negative in the original, the solution was not valid.
00:31:14.800 --> 00:31:20.300
Therefore, there is no solution to this radical equation.
00:31:20.300 --> 00:31:25.500
That concludes this lesson on radical equations and inequalities.
00:31:25.500 --> 00:31:27.000
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