Solve each equation. See Examples 4–6. √(x+2)=1-√(3x+7)

Question

Accepted Answer

Hello, everyone in this video, we're going to be looking at this practice problem where we want to solve for X in the equation, square root of X plus five is equal to two minus the square root of two X plus nine. So in this case, because we're solving for X and X is present in both the left hand side and the right hand side, we're going to try to move X onto one side. And because X is stuck in a square root, we're going to start by squaring both sides. So once I square, both sides, the left hand side becomes X plus five, that is equal to two minus the square root of two X plus nine times two minus the square root of two X plus nine. Now go ahead and multiply these. So two times two is 42 times square root of two x plus nine is too times the square root of two X plus nine. We have negative square root of two X plus nine times two. We got the same result. And then finally negative to a negative square root of two X plus nine times negative square root. Of two X plus nine is positive two x plus nine. From here, I can combine these middle terms to give me negative four times the square root of two X plus nine. And I'll move this over to the left And then I can combine the positive nine with before. So 4-plus 9 is equal to 13. I'm gonna go ahead and move the two X and the 13 to the left hand side by subtracting by two X and 13. So my left hand side becomes X minus two, X is negative X five minus 13 is negative eight And that's equal to four times the square root of two X plus nine. Well, you see we still have X on both sides. So I'm gonna square one more time. So now my left hand side is negative X minus eight times negative X minus eight. That's equal to four times the square root or we can put the exponent here. So four squared and the two X plus nine is free of the square root And four squared is actually 16. So now I can do these multiplication is once again negative X times negative X is positive X squared, negative X times negative eight is positive eight, X negative eight times negative eight, negative eight times negative X is again positive eight X and the negative eight times negative eight is positive 60. For my right hand side is 16 times two, x is 32 x and 16 times nine is 144. And I'll move all my components over to the left. So I'll subtract 32 x and 144. And on the left hand side, you can see I can combine eight X with eight X becomes eight plus eight is 16 X. So I have X squared and 16 X minus 32 X is negative 16 X and positive 64 minus 44 is equal to negative equal to zero. And now you can see that I have the quadratic equation or a quadratic formula where quadratic equation where I have a squared so I can solve X by factoring. So when I factor I want two terms who's first number need to multiply two X squared. So there's only two terms that multiply two X square that would be X with X or negative X and negative X. And in this case, I'm just going to try X positive X. So you can see X times X is X squared. So that gives me my first term. And then I need the last terms to multiply to give me -80. So because I get negative 80, I know that they're going to have opposite signs. And when we think of factors of 80, we can have eight or 10, four and to end 40. Um for the sake I'm just going to try positive for negative 20. So X plus four X minus 20. And we can check this factoring by doing the foil method on our factory ization. So X times X is X squared, X times negative 20 is negative 20 X positive four times X is four X and four times negative 20 is negative 80. And here you can see that The middle terms combined to give me -16 x I start with the -80. So I get back my original equation. So my factoring is good. So now I can solve for X using the factory. So my first factor is X plus four is equal to zero. And my second factor X minus 20 is equal to zero. From here X is equal to negative four and X is equal to positive 20. And you may be thinking those are final solutions. But in order to call them our final solutions, we need to plug them back into our original equation. So our original equation is the one that the problem gave us. So we need to plug it back into the square root of X plus five is equal to two minus the square root of two X plus nine. So I can test my first solution here where X is equal to negative four. So the square root of negative four plus five is equal to two minus the square root of two times negative four plus nine, negative four plus five is one, two minus two times negative four is negative eight plus nine, negative eight plus nine is one. The square root of one is one that's equal to -1, Which means one is equal to one. And this is true. So X equal to negative four is one of our solutions. And then we can do the same for X is equal to 20 and plug it in, I have the square root of 20 plus five is equal to two minus the square root of two times 20 plus nine. So 20 plus five is 25, square root of 25 is equal to two minus square root of two times 20 is 40 plus nine. So 40 plus nine is 49 and the square root of five is 5 equal to 2 - the square root of is seven. So I have five is equal to 2 -7, which is negative five and this is not true. So 20 equal x equal to 20 is not one of our solutions. So that means that x equal to negative four is our only solution for this equation. And if we look at the answer choices, you can see that answer choice B has negative for as the only solution as well. So hopefully this video hoped and see you next time.

Solve each equation. See Examples 4–6. √(x+2)=1-√(3x+7)

Key Concepts

Square Root Equations

Domain Restrictions

Checking for Extraneous Solutions

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