Solve each radical equation in Exercises 88–89. √ (2x-3) + x = 3

Question

Accepted Answer

Welcome back. I'm so glad you're here. We have got to solve for X. And one of those exes is inside square root brackets. So x minus the square root of three, X minus 11 equals three. Now we want to get rid of that square root bracket, we want to free that X. But how do we do it? We have to get it by itself first and I want it to be positive instead of negative. So what I'm going to do is add it to both sides, add the squared of three X -11 to both sides. And so that will cancel out on the left and then I want to bring that three over. So that square root is by itself, that will cancel out here. So now we've got x minus three on the left and the square root of three x minus 11 on the right hand side. Great. Now we can free that three X. We get rid of that square root by squaring it. And what we do to one side, we've got to do to the other. So um yeah, that gives us an x minus three squared and yeah, we've got to free that X by foiling that out. But hey, on the right hand side now we've just got three x minus 11. Cool. So yeah, let's foil this out. We've got x minus three times X minus three still equals three, x minus 11. And our first give us X squared our outsides give us negative three X. Our insides also give us negative three X. And our last negative three times negative three. Give us positive nine still equals three X minus 11. Combine like terms in the middle, that's just this negative three X. And a negative three X. So we have x squared minus six X plus nine equals three X minus 11. Cool. And now lets get everything over on the left side so that the right hand side equals zero. We'll subtract the three X and will add the 11 to both sides. So that will give us X squared minus nine X. Negative six X minus three. X is minus nine X and nine plus 11 is plus equals zero. This is something that we can factor. So what do we factor to be X. Time to get X square. That's usually gonna be X times X. And now looking at this plus 20 we've got a lot of options to get us 20. We can do one times 20 or negative one times negative 20. We've got um two times 10 or the negatives of that. And we've got four times five or the negatives of those. Are we dealing with positives or negatives? Well let's look inside here and this is a negative number that they have to add up to be. So let's do a negative four and negative five to get us that negative nine. So X minus four and x minus five and we can double check this real quick X times X X squared X times negative five negative five X or inside negative four X. And our last will be positive 20 X squared combined. The middle terms minus nine X plus 20. Cool that checks out. So we can now set both of these x minus four equal to zero and we can set our x minus five equals to zero and solve for X as one does so add affordable sides. That will give us X equals four as one of our solutions add five to both sides and that will give us X equals five as our other solution. And cool, we can almost declare victory. We just want to check really quick because we are dealing with a square root. So let's plug our numbers in here. We've got four minus the square root of three times four minus 11 equals three. Okay, so inside that square root, that will be 12 minus 11. The square root of 12 minus 11 carry down the four minus and equals 3. 12 minus 11. That's one that's a positive number. So we're good. Four minus the square root of one equals three. And yeah, that's true because squared of one is 14 minus one does equal three. Let's check the other one. The other one is five, so five minus the square root of three times five minus 11 equals three and we do three times five. That'll be 15 minus 11. Inside the square root bracket five minus that and it's still equals three of 15 minus 11. That's going to be four. So five minus square to four. Cool. That is a positive number. And yet it checks out because square to four is 25 minus two does in fact equal three. We're good with our domain and that means that we can in fact declare victory and our answer choice B yields four and five. Well done, We'll catch you on the next one.

Solve each radical equation in Exercises 88–89. √ (2x-3) + x = 3

Key Concepts

Radical Equations

Isolating the Variable

Extraneous Solutions

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