Solve each equation. See Examples 4–6. 3-√x=√(2√(x)-3)

Question

Accepted Answer

Hello today we're going to be solving for X in the given equation. So what we are given is five minus the square root of X is equal to the square root of eight times the square root of x minus seven. So in order to solve for X, I'm going to need to set the equation equal to zero. And in order to do that I'm gonna want to take out the argument of eight times a squared X -7 out of the square root. In order to get rid of the square root, I can go ahead and square both sides of the given equation. And doing this is going to leave me with five minus the square root of X squared is equal to eight times the square root of x minus seven. Now I'm going to need to simplify the left hand side because we have five minus square root of X squared. In order to simplify that, I'm going to need to foil the quantity five minus square root of X squared is the same thing as saying five minus the square root of X, times five minus the square root of X. So in order to foil this, I'm going to take the first term five and multiply to five and the negative square root of X. And that's going to give me five times five which is five times negative squared X, which is negative five squared of X. Then I'm going to multiply negative square root of X 25 and negative square root of X. And that's going to give me the negative five squared of X and negative square root of X times negative square root of X is going to leave me with positive X. Now here I can go ahead and combine like terms as negative five squared of x minus five squared x is going to give me negative 10 squared of X. So I have 25 minus 10 squared of X plus X. And that's what the left hand side is going to simplify too. So if I substitute this in for the left hand side of the equation, I have 25 minus 10 square root of X plus X is equal to eight square root of x minus seven. Now I can go ahead and add 10 squared of X to both sides of the equation And I can go ahead and add to both sides as well. In order to simplify this, Doing this is going to leave me with 32 plus X is equal to 18 square root of X. Now, in order to set this equal to zero, I'm going to want to remove X from the square root. Once again, we're going to go ahead and square both sides of this equation on the right hand side, I'm going to distribute the export of two to both 18 and the square root of X 18 squared is going to evaluate to give me and square root of X squared is going to give me X. And once again in order to simplify the left hand side, I'm going to need a foil 32 plus X squared what I have here is 32 plus X times plus X. So in order to foil this I'm gonna multiply 32 to 32 which is going to give me 1024 32 2 X. Which is going to give me 32 x. Then I'm gonna multiply X to 32 which is going to give me another 32 X and then multiply X with X which is going to give me X squared, combining the like terms is going to give me 1024 plus 64 X plus X squared. And if I reorder this with respect to the exponents, what I have is X squared plus 64 X plus 1024. So simplifying the left hand side is going to give me this try no meal. So I'm gonna replace 32 plus X squared with the tri no meal and the left hand side is going to become X squared plus 64 X plus 1024. So what I can do now is go ahead and subtract 324 X to both sides of the equation and this is going to leave me with X squared minus X plus 1024 is equal to zero. So currently what I have is a fact herbal trin Amiel as this try no meal is going to factor to the quantities X - times. X -4 is equal to zero. Now what I want to do in order to solve for X is set both of the factors of X equal to zero and solve and doing this is going to give me x minus 256 is equal to zero and x minus four is equal to zero. On the left hand side. I can add 256 to both sides of the equation and that's going to give me X is equal to 256. And on the right hand side if I add four to both sides I get X is equal to four. So currently I have two solutions for X but I need to verify both of the solutions by plugging them back in to the original equation. So in order to verify them I'm gonna take X is equal to 256 and plug it into the original equation. Doing this is going to give me 5 - the square root of 256 is equal to the square root Of eight times the square root of -7. Now the square root of 256 is going to evaluate To Give Me 16. So if I replace the square root of 256 with 16, what I have is 5 -16 is equal to the square root of eight times 16 -7. On the left hand side, 5 -16 is going to give me -11. And on the right hand side eight times 16 is going to give me 128 and 128 -7 is going to give me 121. So I have the square root of 121 and the square root of 121 evaluates to 11. So what I am left with is negative, 11 is equal to 11 which is a false statement. So X is equal to 256 is not going to be a solution. Now I need to check X is equal to four plugging in four to the original equation is going to give me 52 minus the square root of four is equal to the square root of eight times the square root of four minus seven. The square root of four evaluates the two. So replacing the square root of four with two is going to give me five minus two is equal to the square root of eight times two minus seven. On the left hand side, five minus two is going to give me three and on the right hand side eight times two is going to give me 16, and 16 minus seven is going to leave me with nine and the square root of nine evaluates to give me three. So what I have is three is equal to three, which is a true statement. So since the value of X is equal to four, works with the original equation, X is equal to four is going to be the solution to the equation. And with that being said, the answer to this problem is going to be be 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. 3-√x=√(2√(x)-3)

Key Concepts

Radical Equations

Isolating Variables

Extraneous Solutions

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