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

Question

In Exercises 1–38, solve each radical equation.____ _____√x + 2 + √3x + 7 = 1

Accepted Answer

Or the following radical equation solved for one where we have radical Y plus four plus radical four, Y plus 12 is equals to one. Our possible answers are negative three, negative 11 divided by nine, negative 11 5 by nine and Y equals negative three or no solution. Now, to first solve this, I'm gonna try and simplify my radical a little bit Vertical y plus four plus radical four multiplied by Y-plus three by factoring out the four in our second radical can simplify it even further. Y plus four under the radical plus square. The four gives us two multiplied by Y plus three under the radical As equals to one. Finally, I will move my two multiplied by squared of Y plus three over to the right side To get radical y plus four. S equals to 1 -2 radical y plus three. Now we can square both sides to get rid of the square roots square of Y plus four. Squared is just Y plus four. On the right side, we have the foot, You will get 1 -2, y plus three Multiplied by 1 -2 radical y plus three. Foiling you a kid one multiplied by one. It's just one, One multiplied by -2, multiplied by red plus three. Whether they're under the radical gives us minus two radical Y plus three. Similarly, we'll have another one for the inside terms negative two multiplied by a square root of Y plus three. Finally, We have negative two multiplied by a square of by plus three multiplied again by negative two multiplied by Y plus three. That gives us plus four, multiplied by Y plus three. We cannot simplify K Y plus four equals to one -4 radical, y plus three Plus four, y plus 12. We can try to simplify again to get rid of our next radical. We'll move our one over for 12 over and our -4 away what they get Y -4, Y gives us -3, Y 4 -1 - Gives us -9 set that equals to negative four radical Y plus three. We can then square once more to get rid of the radical, which on the left side, you look at nine Y squared Plus 27, Y Plus 27, Y plus Equals to 16 multiplied by Y Plus three. We then get nine Y squared plus 54. Y plus 81 equals to 16, Y plus 48. Now we wanna set everything equals to zero. So we will move our 60 and Y and our 48 back over subtract 16. Y and 48 to get our function nine Y squared Plus 38 Y and finally, 81 -48 gives us 33, All equals to zero. We can now try and factor this out to get are what? With a trial and error? We can determine are factors R nine y plus 11 and Y plus three S equals to zero, we can then set each four factor is equal to zero. To solve for A Y nine, Y plus 11 is equal to zero. Y plus three is equal to zero. We are to solve for Wipes chapter 11, You get Y is equals to negative 11 divided by none. And we get Y is equals to negative three. Now we can test values, we'll plug our values back into our equation to see what we get. So our 1st 1 N 11 divided by nine, you know, square root F N 11 divided by nine plus four plus the square root four multiplied by n of 11 divided by nine Plus 12 should equals one. If we were to simplify this and solve, We would end up with 13 divided by three as equals to one. That is not true. So negative 11 divided by nine is not a possible solution. Must also check -3. We have the square root of negative three plus four Plus the square root four multiplied by negative three plus 12 And should equals one nega three plus four gives us a square root of one than we have negative 12 plus 12, which will be zero squared of one Plus zero equals to one Which simplifies the one equals 1. We can then confirm the answer to our problem because y equals negative three which corresponds to answer A OK. I hope to help you solve the problem. Thank you for watching. Goodbye.

In Exercises 1–38, solve each radical equation. _√x + 2 + √3x + 7 = 1

Key Concepts

Radical Equations

Isolating the Radical

Extraneous Solutions

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