Solve each equation. See Examples 4–6. √x+2=√(4+7√x)

Question

Accepted Answer

Hello. Today we're going to be sorry for X in the given equation. So what we are given is the square root of X plus four is equal to the square root of 16 plus nine times the square root of X. Now, in order to solve for X, I'm going to need to set this equation equal to zero. And in order to start that process I'm going to want to remove 16 plus nine squared of X out of the square root. In order to get rid of the square root, I can go ahead and square both sides of the equation and doing this is going to leave me with the square root of X plus four, quantity squared is equal to 16 plus nine times the square root of X. Now on the left hand side, I'm going to need to simplify the square root of X plus four squared. And in order to do that, I'm going to need to foil the quantity. The square root of X plus four squared is the same thing as saying the square root of X plus four times the square root of X plus four. And in order to foil this, I'm going to take the square root of X. Multiply to the square root of X. Which is just going to leave me with X. Then I'm gonna multiply the square root of X with four which is going to give me positive four square root of X. Then I'm going to multiply four with the square root of X. That's going to give me four times the square root of X. And then finally multiply four with four, which is going to give me 16. I can go ahead and combine both of the four square root of X to give me eight square root of X. And that's going to give me X plus eight square root of X plus 16. So this is what the left hand side is going to simplify to. So if I put this into the equation, what I have is X plus eight times the square root of X plus 16 is equal to 16 plus nine squared of X. Now, in order to set this equation equal to zero, I can subtract 16 from both sides of the equation And I can subtract nine squared of X from both sides of the equation. This is going to leave me with X minus the square root of X is equal to zero. And what I want to do now is go ahead and add the square root of X to the right hand side to make everything positive. So adding the square X to both sides is going to give me X is equal to the square root of X. Now, what I want to do is remove the square root of X or remove X out of the square root. And in order to do that, I'm going to square both sides of the equation. Doing this is going to give me X squared is equal to X. Now again we want to set this equation equal to zero. So I'm going to subtract X to both sides of the equation. Doing so is going to give me x squared minus X is equal to zero. And what I could do here is factor out an X from both of the terms, factoring out an X is going to leave me with x times the quantity X -1 is equal to zero. Now in order to solve for X, I'm going to take both of the factors of X. Set it equal to zero and solve So doing that is going to give me X is equal to zero and X minus one is equal to zero. X is equal to zero is a potential solution. And on the right hand side I can add one to both sides to give me X is equal to one. Now that I have both of the solutions for X, I need to go ahead and verify these solutions and in order to verify the solutions, I'm going to take them and plug them back into the original quantity or the original equation. So let's go ahead and test X is equal to zero, plugging in X is equal to zero is going to give me the square root of zero plus four is equal to the square root of 16 plus nine times the square root of zero. Now the square root of zero is just going to give me zero. So on the left hand side, zero plus four is going to give me four. And on the right hand side zero times nine is going to give me zero. So what I am left with is the square root of 16. The square root of 16 evaluates to give me four. So what I have is for is equal to four which is a true statement. So X is equal to zero is a solution. Now we want to check, X is equal to one. Plugging in one into the original equation is going to give me the square root of one plus four is equal to the square root of 16 plus nine times the square root of one. The square root of one is just going to be one. So on the left hand side I have one plus four and on the right hand side I have 16 plus nine. One plus four is going to give me five and 16 plus nine is going to give me 25. So I have five is equal to the square root of 25. And the square root of 25 is going to evaluate to give me five. So what I am left with is five is equal to five which is a true statement. So since both of the solutions worked, that means X is going to equal to zero and one and this is going to be the solutions of X to the original equation. And with that being said, the answer to this problem is going to be C. So I hope this video helps you in understanding how to approach this problem, and I'll go ahead and see you all in the next video.

Solve each equation. See Examples 4–6. √x+2=√(4+7√x)

Key Concepts

Square Roots

Isolating Variables

Squaring Both Sides

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