Unit 3:Solving Radical Equations!

This is an example of a problem, following the rules of solving radical equations.

Step 1: √x+2=x+4

  • My first step was to create my original equation and write it down.
  • Step 2: √(x+2)^2=(x+4)^2

  • My second step was to square both sides in order to get rid of the square root symbol around the x+2. Once I squared the square root of x+2, the square root and the square cancelled each other out so I was left with x+2. I then moved on to squaring the x+4 and when I did I was left with (x+4)(x+4).
  • Step 3: x+2=x^2+4x+4x+16

  • Next, I moved on to factoring the (x+4)(x+4) and I did so by Foiling it. When using the FOIL method, I had to first multiply the first terms which were the x and the x. When I did that, I got x2. My next step was to multiply the outer terms which were the x and 4. When I multiplied those two together, I was left with 4x. Next, I had to multiply the inner terms together which were 4 and x. When I multiplied those terms together I was left with 4x. Lastly, I had to multiply the last terms together which were 4 and 4. When I did, I got the answer of 16. After Foiling each term, I put them together in numerical order of the exponents and I ended off with the equation shown above.
  • Step 4: x+2=x^2+8x+16

  • I then did the minor step of combining like terms. I found the two terms that had an x and a number in front of it and when I did I saw that the two terms that had that were the two 4x’s. I combined them by addition, I added the two 4’s together and got 8 and then I carried the x down. When I did that I got 8x and substituted that in for the 4x+4x. I was then left with the equation shown above.
  • Step 5: x+2-x-2=x^2+8x+16-x-2

  • After that, I knew that I had to bring everything to the same side so I found the term with the biggest exponent and took the proper steps in order to move everything to that side. I knew that I had to move the x+2 to the other side so I did so by doing the opposite operation to both terms. I saw that the x was positive so I moved it to the other side by subtracting it. When I did that it cancelled out on that side and moved over to the other. I then saw that the 2 was being added so I moved it over by subtracting it. When I subtracted it, it cancelled out from that side and moved over.
  • Step 6: x^2+8x-1x+16-2=0 → x^2+7x+14=0

  • After completing the previous step I was left with the equation shown above. With this equation I had to start to simplifying the like terms so I did so by subtracting the 1x from the 8x which left me with 7x and then I subtracted the 2 from the 16 and got 14. After getting these numbers I put them all together as the second equation shown and then I set that to 0.
  • Step 7: mn/ m+n

  • Next, I realized that I had three terms set to 0 so I knew that I then had to use the mn/m+n method to factor it. In order to factor it I had to figure out what two numbers could multiply together and get the last term which is 14 and add together to get the middle term which is 7. After thinking of all of the factors of 14, I came to the conclusion that no set of numbers could perform both of those operations so I marked it as cannot be factored.
  • Step 8: Quadratic Formula

  • After coming to the conclusion that it could not be factored, I used my knowledge and came to the idea of using the quadratic formula. I first wrote out the quadratic formula which is -b ± √b^2-4ac÷2a. After writing it down I knew that in order to substitute my terms in, I had to first identify my variables. I set a=1, b=7, and c=14 due to the order in which they were in the equation.
  • Step 9: -7 ± √7^2-4(1)(14)÷2(1)

  • My next step was to use the numerical values of the variables in order to substitute them in and simply it down. I substituted the 7 in for each b term, the 1 in for every a term, and th 14 in for every c term in the equation.
  • Step 10: -7 ±√49-4(14)÷2

  • I then started to multiply the terms together in order to simplify it even more. When I multiplied the 7^2 I got 49 so I replaced it in the equation and then I multiplied the 4 and the one together in order to get the 4 and replace it in the equation. I was then able to the denominator and multiply the 2 and the 1 in order to get the 2 to replace it with.
  • Step 11: -7 ±√ 49-56 ÷2

  • I then moved on to the minor step of multiplying the 4 and the 14 together and when I did I got 56, so I added that in and got the equation above.
  • Step 12: -7 ±√ -7 ÷ 2

  • After completing the step above, I subtracted the terms inside of the radical, which were 49 and 56. When I subtracted the 56 from the 49 I was left with -7, so I replaced them in the equation and added the -7 underneath the radical sign.
  • Step 13: -7 ±i√ 7÷ 2

  • I then moved on to trying to further simplify it and when I did, I realized that there was a negative inside of the radical so I knew that I had to get rid of it. I used my knowledge of the topic in order to take out the negative sign and add an “i” in front of the radical. I knew that I had to do this because of the rules of imaginary numbers.
  • Step 14: -7± 2.65÷ 2→ -7 + 2.65÷2 → -4.35÷2

  • My next step was to find the square root of 7, so I did by inputting it into my calculator. When I did that I found that the answer was 2.65 so with that I substituted it into the equation and got the equation above. After getting that I realized that I had to start to follow the rules of the sign so I did so by adding 2.65 to the -7. When I did that I ended up having to subtract them, which left me with -4.35. After getting that I had to carry the 2 over in the denominator.
  • Step 15: -2.18i

  • Next, I had to simplify it down more by dividing -4.35 by 2. When I did that I was left with -2.18 and then I had to carry the imaginary number(i) back down. When I did that I added it to the end of my term which left me with the number and variable shown above.
  • Step 16: -7- 2.65 ÷2 → - 9.65÷2

  • After completing the addition part of the ± sign I had to complete the subtraction part of it. I did this by subtracting -2.65 from the -7. Due to the rules of signs, because of the fact that both numbers have a subtraction sign in front of it I had to add them and then carry the sign. When I added the 7 ad 2.65 I was left with -9.65. I then left that over the 2 in the denominator.
  • Step 17: -4.83i

  • My last step of the problem was to complete the division for the equation above and carry down the imaginary number(i). When I did that I was left with -4.83 from doing -9.65/2 and then I carried down my “i”.