In Exercises 1–38, solve each radical equation.
____
2√x - 3 + 4...

Question

In Exercises 1–38, solve each radical equation.
____ 
2√x - 3 + 4 = x + 1

Accepted Answer

Welcome back. I am so glad you're here. We are asked in the following given radical equation to solve for X. Our given radical equation is two multiplied by the square root of X minus seven. Then outside of the radical plus five equals X minus two. Our answer choices are answer choice A, the solution set negative 11 negative seven answer choice B the solution set negative 11 7 answer choice C the solution set negative 7-Eleven and answer choice D the solution set 7-Eleven. All right. In order to solve for X, we want to have X all by itself on one side of the equal sign and all the regular numbers on the other side. However, in this case, we've got two XS and they're on opposite sides of the equal sign and one of them is trapped underneath a radical. So the first thing we need to do is actually rescue that X that's trapped inside of the radical. So we want to isolate the radical and then get rid of the radical. So we will be following the reverse of the order of operations. We recall from previous lessons that acronym peda for parentheses exponents multiplication and division addition and subtraction. We'll be starting off looking for addition and subtraction that's not tied directly to, it's not directly inside of the radical. So on the left, we've got this plus five five is, is being added, we want to do the opposite of addition, which is subtraction. So we will subtract five from both sides of the equation. On the left, positive five minus five is zero. So we'll just have two multiplied by the square rut of X minus seven on the left, that's equal to, on the right. Still X but now we have negative two minus five, which is going to be a negative seven. So we're getting that radical isolated. There's no more addition or subtraction over there. Now there's multiplication that two is being multiplied by the radical. So we're going to do the opposite of multiplication. We'll divide and we'll divide both sides by two. Because what we do to one side, we do to the other on the left two divided by two is one. So we just have the square root of X minus seven on the left, that's equal to, I'm going to keep this as a fraction X minus seven in the numerator and then divided by two in the denominator. So now we have the radical isolated. Yay. Now to rescue that X from the radical when we're taking the square root, the way to undo that is to raise it to the exponent two because the square root of something squared is going to be what's inside of the radical and what we do to one side, we do to the other. So we'll be squaring the right hand side as well. On the left, the square root of X minus seven squared is just X minus seven. Yay. Now, on the right, we do need to take, we have the quantity of X minus seven in the numerator. We will be squaring that and that's divided by two squared in the denominator which is four. So let's do the work for the quantity of X minus seven multiplied by the quantity of X minus seven. Or in other words, we're squaring X minus seven, our numerator. So taking our first X multiplied by X that will give us X squared our outsides X multiplied by negative seven. That will be a negative seven X. Our insides negative seven multiplied by X. Sorry, grab the wrong color, negative seven multiplied by X. That's going to be again negative seven X and our last negative seven multiplied by negative seven is a positive 49. Combining our like terms in the middle, negative seven X minus seven X will be a negative 14 X and bring down the X squared in front and the plus 49 afterward. So that is our new numerator. So we have X minus seven on the left is equal to in the numerator, X squared minus 14 X plus 49 all divided by four. OK. Now, our plan to get all of the XS together we're dealing with an X squared this time. So when we've got a quadratic, we want to try to set one side equal to zero so that we can hopefully factor this quadratic and solve for X that way. So to get the quadratic by itself and zero on the other side, let's first get rid of our fraction. So we're going to multiply both sides of this equation by the denominator by four, which will get rid of our fraction on the right. So on the left, we will be distributing this four to both terms inside of the parentheses, we will be distributing four to X and four to negative 74, multiplied by X is four X and four multiplied by negative seven is a negative 28. And that's equal to on the right, the four is cancel out. So this is just X squared minus 14 X plus 49. Now to get the quadratic by itself, we will subtract four X from the left and from the right. What we do to one side, we do to the other because on the left four X minus four X is zero. And then we have that negative 28 we will add 28 to both sides because negative 28 plus 28 is zero. So the left hand side is equal to zero. Yay. On the right, we still have X squared and the negative 14 X minus four X is negative 18 X and positive plus 28 is plus 77. So I'm going to write this up here where we'll have a little more room X squared minus 18 X plus 77 is equal to zero. Now, we can attempt to factor this to get solving for X. So we can do the reverse of the foil process. Setting up two open sets of parentheses and this is equal to zero. We start by asking ourselves what multiplied by what gives us X squared. Those will be our first and that would be X multiplied by X. Next, we will ask ourselves what multiplied by what equals a positive 77. Those will be our lasts. But keeping in mind when we take our outsides and we add those to our insides, they need to add up to a negative 18 X. So in other words, what two factors of positive 77 add up to a negative 18, that's going to be negative 11 and negative seven. If you need to pause the video and foil that out, just to check and make sure we factored correctly, go for it. Otherwise I am going to continue. So solving for X, we now know that if either of those factors were zero, this statement would be true. So to solve for X, we set both of those factors equal to zero, we set X minus 11 equal to zero and X minus seven equal to zero for X minus 11 equals zero to get that X by itself, we have that 11 being subtracted. We add it to both sides of the equation. So a negative 11 plus 11 cancels out to zero. And we just have X on the left zero plus 11 is positive 11. So X equals 11. Now we can test X minus 11 equals zero. We can add seven to both sides. And on the left negative seven plus seven is zero. So we just have X that's equal to, on the right zero plus seven is a positive seven. So we have our two solutions for X X is equal to 11 and X is equal to seven. We look at our answer choices and this matches with answer choice. D well done. We'll catch you on the next one.

In Exercises 1–38, solve each radical equation. ____ 2√x - 3 + 4 = x + 1

Key Concepts

Radical Equations

Isolating Variables

Extraneous Solutions

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