https://padlet.com/kspangler/quadratic Write your equation as the "Title." Then, where it says "write something," explain how you know your equation belongs in that column and what your process was for writing that equation. en-us 2016-10-24 16:42:13 UTC 2017-05-17 11:04:32 UTC hello@padlet.com https://padlet.com/kspangler/quadratic/wish/132816972 y=x^2]]> 2016-10-24 18:31:59 UTC https://padlet.com/kspangler/quadratic/wish/132816972 michaelm1384 https://padlet.com/kspangler/quadratic/wish/132816982
This equation works because (x+3) x (x-3) does not create a positive 9 in the intercept. The equation can only be solved by quadratic formula which reveals a -9 that is being square rooted. A square root cannot contain a negative number. ]]> 2016-10-24 18:32:01 UTC https://padlet.com/kspangler/quadratic/wish/132816982 gradys1345 https://padlet.com/kspangler/quadratic/wish/132817290 y=x^2x+1
The equation would make it so there is a zero in the radical making the answer -1]]> 2016-10-24 18:32:46 UTC https://padlet.com/kspangler/quadratic/wish/132817290 houstonr7262 https://padlet.com/kspangler/quadratic/wish/132817365 Radical 0 is zero, so there is no need for the +- that makes it a single solution]]> 2016-10-24 18:32:58 UTC https://padlet.com/kspangler/quadratic/wish/132817365 brennanp0703 https://padlet.com/kspangler/quadratic/wish/132817456 2016-10-24 18:33:11 UTC https://padlet.com/kspangler/quadratic/wish/132817456 madelyna7189 https://padlet.com/kspangler/quadratic/wish/132817821 The answer would be
 (x-4)^2 which would leave only 1 answer, 4.]]> 2016-10-24 18:34:06 UTC https://padlet.com/kspangler/quadratic/wish/132817821 nicholasd2080 https://padlet.com/kspangler/quadratic/wish/132817838 0=x^+4x+4
It would put a radical over the zero, making it only one solution.  

]]> 2016-10-24 18:34:09 UTC https://padlet.com/kspangler/quadratic/wish/132817838 mitchellSains https://padlet.com/kspangler/quadratic/wish/132817982 2016-10-24 18:34:29 UTC https://padlet.com/kspangler/quadratic/wish/132817982 trac_starz https://padlet.com/kspangler/quadratic/wish/132817989 One solution because when factored it only has 3 as the solution no other #. (x+3)^2]]> 2016-10-24 18:34:31 UTC https://padlet.com/kspangler/quadratic/wish/132817989 bradonw3792 https://padlet.com/kspangler/quadratic/wish/132818018 I know this has one solution because y is set to zero making it only 0]]> 2016-10-24 18:34:37 UTC https://padlet.com/kspangler/quadratic/wish/132818018 madelyna7189 https://padlet.com/kspangler/quadratic/wish/132818087 Since this equation is not a perfect square, there has to be 2 solutions when you factor it.]]> 2016-10-24 18:34:49 UTC https://padlet.com/kspangler/quadratic/wish/132818087 danielm2984 https://padlet.com/kspangler/quadratic/wish/132818228 2016-10-24 18:35:09 UTC https://padlet.com/kspangler/quadratic/wish/132818228 nicholasm7203 https://padlet.com/kspangler/quadratic/wish/132818231 2016-10-24 18:35:09 UTC https://padlet.com/kspangler/quadratic/wish/132818231 bradonw3792 https://padlet.com/kspangler/quadratic/wish/132818344 2016-10-24 18:35:26 UTC https://padlet.com/kspangler/quadratic/wish/132818344 brennanp0703 https://padlet.com/kspangler/quadratic/wish/132818368 When you factor 6 and 9 which makes (x-3).

]]> 2016-10-24 18:35:29 UTC https://padlet.com/kspangler/quadratic/wish/132818368 colind0687 https://padlet.com/kspangler/quadratic/wish/132818478 4 times 3 is 12 and 4 plus 3 equals 7 so it factors to -3 and -4.]]> 2016-10-24 18:35:43 UTC https://padlet.com/kspangler/quadratic/wish/132818478 nicholasd2080 https://padlet.com/kspangler/quadratic/wish/132818629 y=x^-4x+4
the 4 factors into -2 which then makes it (x-2) ]]> 2016-10-24 18:36:06 UTC https://padlet.com/kspangler/quadratic/wish/132818629 mitchellSains https://padlet.com/kspangler/quadratic/wish/132818639 it would put a radical over 0 which is zero, making it only one solution x = -1]]> 2016-10-24 18:36:07 UTC https://padlet.com/kspangler/quadratic/wish/132818639 seand7054 https://padlet.com/kspangler/quadratic/wish/132818693 it would radical zero making it have only one solution: x = -4]]> 2016-10-24 18:36:15 UTC https://padlet.com/kspangler/quadratic/wish/132818693 sophiam3244 https://padlet.com/kspangler/quadratic/wish/132818813 y=3x^2 -24x+48
Once you get rid of a (3) , you get x^2 -8x+16 and you can factor that to get 3(x-4)^2 which comes to two solutions of 6 and 2]]> 2016-10-24 18:36:30 UTC https://padlet.com/kspangler/quadratic/wish/132818813 vanessal3761 https://padlet.com/kspangler/quadratic/wish/132818936 factors to x=-5 and x=-3 b/c  5  and 3 multiply to 15 and add up to 8]]> 2016-10-24 18:36:43 UTC https://padlet.com/kspangler/quadratic/wish/132818936 nikolash2579 https://padlet.com/kspangler/quadratic/wish/132818960 when factored the solution is -4 ]]> 2016-10-24 18:36:45 UTC https://padlet.com/kspangler/quadratic/wish/132818960 kevinl3942 https://padlet.com/kspangler/quadratic/wish/132818989 The factors are 2 and 3 because 2 and 3 multiply to 6 and add to 5]]> 2016-10-24 18:36:48 UTC https://padlet.com/kspangler/quadratic/wish/132818989 https://padlet.com/kspangler/quadratic/wish/132819052 x^2+3x+4=32

There is only one answer.]]> 2016-10-24 18:36:54 UTC https://padlet.com/kspangler/quadratic/wish/132819052 houstonr7262 https://padlet.com/kspangler/quadratic/wish/132819088 the equation becomes 
x^2-2x+2x-4 which factors down to (x+2)(x-2). The answers are +2 and -2]]> 2016-10-24 18:36:59 UTC https://padlet.com/kspangler/quadratic/wish/132819088 bobbyg6791 https://padlet.com/kspangler/quadratic/wish/132819111 This only has 1 solution because it comes out to (x+3)^2]]> 2016-10-24 18:37:02 UTC https://padlet.com/kspangler/quadratic/wish/132819111 nicholasd2080 https://padlet.com/kspangler/quadratic/wish/132819120 y=x^+4x+9
the 9 does not factor into 4 but it can still be solved by the quadratic function. ]]> 2016-10-24 18:37:03 UTC https://padlet.com/kspangler/quadratic/wish/132819120 nicholasm7203 https://padlet.com/kspangler/quadratic/wish/132819212 2016-10-24 18:37:15 UTC https://padlet.com/kspangler/quadratic/wish/132819212 alexisn1862 https://padlet.com/kspangler/quadratic/wish/132819529 You can factor to end up with (x+10) and (x+2) which comes to two solutions.]]> 2016-10-24 18:37:53 UTC https://padlet.com/kspangler/quadratic/wish/132819529 christopherm686 https://padlet.com/kspangler/quadratic/wish/132819584 x=2 because 4-4=0]]> 2016-10-24 18:38:00 UTC https://padlet.com/kspangler/quadratic/wish/132819584 madelyna7189 https://padlet.com/kspangler/quadratic/wish/132819790 There is no number that multiply to 2 & add to -4 so you would need to use the quadratic formula for it.]]> 2016-10-24 18:38:30 UTC https://padlet.com/kspangler/quadratic/wish/132819790 brennanp0703 https://padlet.com/kspangler/quadratic/wish/132819865 The x and the 100 cannot be factor because it equals out to itself. 

]]> 2016-10-24 18:38:42 UTC https://padlet.com/kspangler/quadratic/wish/132819865 maryamj2739 https://padlet.com/kspangler/quadratic/wish/132819964 when factored x=9 x=10 because (x-10)(x-9)]]> 2016-10-24 18:38:57 UTC https://padlet.com/kspangler/quadratic/wish/132819964 colind0687 https://padlet.com/kspangler/quadratic/wish/132820105 this factors to 1 and 1 so you would do x+1=0 and it's so -1 is the solution.]]> 2016-10-24 18:39:12 UTC https://padlet.com/kspangler/quadratic/wish/132820105 trac_starz https://padlet.com/kspangler/quadratic/wish/132820257 When factored, the two solutions would be -9 and 4 because they add up to -5 and multiply to -36]]> 2016-10-24 18:39:35 UTC https://padlet.com/kspangler/quadratic/wish/132820257 seand7054 https://padlet.com/kspangler/quadratic/wish/132820317 x^2 + 6x __+9__ =  -16
(x + 3)^2 = -16
x + 3 = +/- 16
rad both sides and the answers will be 13 and -19]]> 2016-10-24 18:39:43 UTC https://padlet.com/kspangler/quadratic/wish/132820317 mitchellSains https://padlet.com/kspangler/quadratic/wish/132820557 when factored the solutions are -3,-5]]> 2016-10-24 18:40:15 UTC https://padlet.com/kspangler/quadratic/wish/132820557 https://padlet.com/kspangler/quadratic/wish/132820774 When solved using the quadratic formula, the discriminant becomes a negative, and the whole equation/fraction cannot be simplified.]]> 2016-10-24 18:40:43 UTC https://padlet.com/kspangler/quadratic/wish/132820774 madelyna7189 https://padlet.com/kspangler/quadratic/wish/132820837 This would have imaginary solutions because the number in the square root is negative and greater than the numbers out of it, which would make x negative.]]> 2016-10-24 18:40:52 UTC https://padlet.com/kspangler/quadratic/wish/132820837 sophiam3244 https://padlet.com/kspangler/quadratic/wish/132820852 Using the quadratic formula, you can solve to find that the solution is 1 ]]> 2016-10-24 18:40:54 UTC https://padlet.com/kspangler/quadratic/wish/132820852 https://padlet.com/kspangler/quadratic/wish/132820894 2016-10-24 18:41:01 UTC https://padlet.com/kspangler/quadratic/wish/132820894 houstonr7262 https://padlet.com/kspangler/quadratic/wish/132820975 This is way too difficult to factor, and no simple integers add to 27 and multiply to 30. It would be better to plug them in to the Quadratic formula. ]]> 2016-10-24 18:41:11 UTC https://padlet.com/kspangler/quadratic/wish/132820975 christopherm686 https://padlet.com/kspangler/quadratic/wish/132821102 x=1 because 1+1+2]]> 2016-10-24 18:41:23 UTC https://padlet.com/kspangler/quadratic/wish/132821102 gracek7295 https://padlet.com/kspangler/quadratic/wish/132821110 it can factor evenly to (x+4) and (x+2). Set both of these equations equal to zero and there are 2 real numbers as solutions]]> 2016-10-24 18:41:24 UTC https://padlet.com/kspangler/quadratic/wish/132821110 mariam3926 https://padlet.com/kspangler/quadratic/wish/132821286 it equals -23]]> 2016-10-24 18:41:46 UTC https://padlet.com/kspangler/quadratic/wish/132821286 maryamj2739 https://padlet.com/kspangler/quadratic/wish/132821426 the negative sign under the square root becomes an imaginary number
x=-9i x=9i ]]> 2016-10-24 18:42:02 UTC https://padlet.com/kspangler/quadratic/wish/132821426 priscillab7507 https://padlet.com/kspangler/quadratic/wish/132821459 When you factor the equation, you end up with the factors (x-4) and (x-3). This is then simplified down to two real solutions. ]]> 2016-10-24 18:42:06 UTC https://padlet.com/kspangler/quadratic/wish/132821459 kevinc3655 https://padlet.com/kspangler/quadratic/wish/132821515 when the 2 is subtracted to both sides it equals x=1 which is one solution]]> 2016-10-24 18:42:15 UTC https://padlet.com/kspangler/quadratic/wish/132821515 https://padlet.com/kspangler/quadratic/wish/132821709 Radical zero doesn't need plus or minus in front and factoring would get x=5]]> 2016-10-24 18:42:41 UTC https://padlet.com/kspangler/quadratic/wish/132821709 colind0687 https://padlet.com/kspangler/quadratic/wish/132821744 if you try to factor this it won't so you would use the quadratic formula.]]> 2016-10-24 18:42:45 UTC https://padlet.com/kspangler/quadratic/wish/132821744 sophiam3244 https://padlet.com/kspangler/quadratic/wish/132821965 3x^2+5=41 Since this cannot be factored, you can solve by square roots to get the solution of 2 rad 3 and -2 rad 3]]> 2016-10-24 18:43:15 UTC https://padlet.com/kspangler/quadratic/wish/132821965 nicholasd2080 https://padlet.com/kspangler/quadratic/wish/132822000 When put into the quadratic function it makes the number inside of the radical negative. ]]> 2016-10-24 18:43:20 UTC https://padlet.com/kspangler/quadratic/wish/132822000 mitchellSains https://padlet.com/kspangler/quadratic/wish/132822085 the 10 doesnt factor in 5, but it can still be solved by using the quadratic formula.]]> 2016-10-24 18:43:31 UTC https://padlet.com/kspangler/quadratic/wish/132822085 houstonr7262 https://padlet.com/kspangler/quadratic/wish/132822111 this becomes 4x^2=-9, and then you need to go radical, which gets imaginary numbers]]> 2016-10-24 18:43:34 UTC https://padlet.com/kspangler/quadratic/wish/132822111 https://padlet.com/kspangler/quadratic/wish/132822123 y=1
This answer was only 1 because 
-Imani moreen]]> 2016-10-24 18:43:36 UTC https://padlet.com/kspangler/quadratic/wish/132822123 nikolash2579 https://padlet.com/kspangler/quadratic/wish/132822157 when factored the solutions are -4,-2]]> 2016-10-24 18:43:40 UTC https://padlet.com/kspangler/quadratic/wish/132822157 https://padlet.com/kspangler/quadratic/wish/132822159 This has only one solution because when factored, it's a perfect square (x+3)^2. Therefore, the quadratic should have only one solution.]]> 2016-10-24 18:43:40 UTC https://padlet.com/kspangler/quadratic/wish/132822159 nicholasm7203 https://padlet.com/kspangler/quadratic/wish/132822364 This can not be factored but can b solved by the quadratic equation]]> 2016-10-24 18:44:07 UTC https://padlet.com/kspangler/quadratic/wish/132822364 trac_starz https://padlet.com/kspangler/quadratic/wish/132822462 This can't be factored. You can factor this be plugging the a,b, and c values into the quadratic formula to solve without factoring.]]> 2016-10-24 18:44:15 UTC https://padlet.com/kspangler/quadratic/wish/132822462 briannar7340 https://padlet.com/kspangler/quadratic/wish/132822790 You can factor this to make (x+5)(x+2) which means the solutions are -5 and -2, which are both real numbers

]]> 2016-10-24 18:45:00 UTC https://padlet.com/kspangler/quadratic/wish/132822790 danielm2984 https://padlet.com/kspangler/quadratic/wish/132822841 2016-10-24 18:45:08 UTC https://padlet.com/kspangler/quadratic/wish/132822841 christopherm686 https://padlet.com/kspangler/quadratic/wish/132822895 x=-2 and when you solve it becomes -40=0 ]]> 2016-10-24 18:45:14 UTC https://padlet.com/kspangler/quadratic/wish/132822895 sophiam3244 https://padlet.com/kspangler/quadratic/wish/132823127 Once you solve it through the use of the quadratic formula, you end up with the solutions of x=1 +3i and x= 1-3i]]> 2016-10-24 18:45:46 UTC https://padlet.com/kspangler/quadratic/wish/132823127 gracek7295 https://padlet.com/kspangler/quadratic/wish/132823185 there is a zero in the radical so it would solve out perfectly with only one solution. It is also a perfect square]]> 2016-10-24 18:45:54 UTC https://padlet.com/kspangler/quadratic/wish/132823185 bradonw3792 https://padlet.com/kspangler/quadratic/wish/132823492 2016-10-24 18:46:39 UTC https://padlet.com/kspangler/quadratic/wish/132823492 alexisn1862 https://padlet.com/kspangler/quadratic/wish/132823687 This equation can't be factored because 3 doesn't factor into 10. You can solve by using the quadratic formula]]> 2016-10-24 18:47:07 UTC https://padlet.com/kspangler/quadratic/wish/132823687 christopherm686 https://padlet.com/kspangler/quadratic/wish/132823744 since you cant factor this you use the quadratic formula to solve for the x values]]> 2016-10-24 18:47:17 UTC https://padlet.com/kspangler/quadratic/wish/132823744 vanessal3761 https://padlet.com/kspangler/quadratic/wish/132823848 the solution is -4 b/c when it factors you get a perfect square. (x+4)^2]]> 2016-10-24 18:47:32 UTC https://padlet.com/kspangler/quadratic/wish/132823848 gracek7295 https://padlet.com/kspangler/quadratic/wish/132823870 there will be a negative number inside of
 the radical making it an imaginary number]]> 2016-10-24 18:47:35 UTC https://padlet.com/kspangler/quadratic/wish/132823870 danielm2984 https://padlet.com/kspangler/quadratic/wish/132823871 It can only be factored to end up with one solution because it factors to X+1 and x-1]]> 2016-10-24 18:47:35 UTC https://padlet.com/kspangler/quadratic/wish/132823871 trac_starz https://padlet.com/kspangler/quadratic/wish/132823898 When factored the x will become -6 and 6. This doesn't multiply to 36. When placed in a radical it will have an imaginary # (or a negative)]]> 2016-10-24 18:47:39 UTC https://padlet.com/kspangler/quadratic/wish/132823898 colind0687 https://padlet.com/kspangler/quadratic/wish/132824249 Subtracting 9 would get x^2=-9 and then square both sides and it would be squaring a negative]]> 2016-10-24 18:48:40 UTC https://padlet.com/kspangler/quadratic/wish/132824249 maryamj2739 https://padlet.com/kspangler/quadratic/wish/132824256 solve by using the quadratic formila]]> 2016-10-24 18:48:41 UTC https://padlet.com/kspangler/quadratic/wish/132824256 priscillab7507 https://padlet.com/kspangler/quadratic/wish/132824315 When solved, the solution is 1.]]> 2016-10-24 18:48:47 UTC https://padlet.com/kspangler/quadratic/wish/132824315 kevinc3655 https://padlet.com/kspangler/quadratic/wish/132824356 the 4 is added to both side and then squared, it ends up being x=+-]]> 2016-10-24 18:48:54 UTC https://padlet.com/kspangler/quadratic/wish/132824356 seand7054 https://padlet.com/kspangler/quadratic/wish/132824531 can only be solved with quadratic formula]]> 2016-10-24 18:49:26 UTC https://padlet.com/kspangler/quadratic/wish/132824531 mitchellSains https://padlet.com/kspangler/quadratic/wish/132824625 when plugged into the quadratic formula it makes the number inside the radical a negative

]]> 2016-10-24 18:49:39 UTC https://padlet.com/kspangler/quadratic/wish/132824625 https://padlet.com/kspangler/quadratic/wish/132824785 y=x^
-evan fredericks]]> 2016-10-24 18:50:11 UTC https://padlet.com/kspangler/quadratic/wish/132824785 bobbyg6791 https://padlet.com/kspangler/quadratic/wish/132824873 there would be 2 solutions because it would end up being (x+5)(x+4). ]]> 2016-10-24 18:50:30 UTC https://padlet.com/kspangler/quadratic/wish/132824873 gradys1345 https://padlet.com/kspangler/quadratic/wish/132824876 y=x^2 4x+4

]]> 2016-10-24 18:50:31 UTC https://padlet.com/kspangler/quadratic/wish/132824876 bobbyg6791 https://padlet.com/kspangler/quadratic/wish/132824987 this equation could not be found by factoring because there is no number that multiplies to 6 and adds to 12. ]]> 2016-10-24 18:50:52 UTC https://padlet.com/kspangler/quadratic/wish/132824987 https://padlet.com/kspangler/quadratic/wish/132825040 2016-10-24 18:51:02 UTC https://padlet.com/kspangler/quadratic/wish/132825040 https://padlet.com/kspangler/quadratic/wish/132825103 y=x^+4x-4
b factors into 2 and then changes to x+2
-evan fredericks]]> 2016-10-24 18:51:14 UTC https://padlet.com/kspangler/quadratic/wish/132825103 https://padlet.com/kspangler/quadratic/wish/132825109 y=x^2-5x-36
Get solutions of +9 and -4]]> 2016-10-24 18:51:15 UTC https://padlet.com/kspangler/quadratic/wish/132825109 kevinc3655 https://padlet.com/kspangler/quadratic/wish/132825137 when the 18 is subtracted to both sides it amkes the other side -16 there fore becomes an imaginary number]]> 2016-10-24 18:51:21 UTC https://padlet.com/kspangler/quadratic/wish/132825137 seand7054 https://padlet.com/kspangler/quadratic/wish/132825144 can only have imaginary answers because the radical will be negative.]]> 2016-10-24 18:51:24 UTC https://padlet.com/kspangler/quadratic/wish/132825144 bobbyg6791 https://padlet.com/kspangler/quadratic/wish/132825195 2016-10-24 18:51:33 UTC https://padlet.com/kspangler/quadratic/wish/132825195 gradys1345 https://padlet.com/kspangler/quadratic/wish/132825218 x^2 +6x + 9]]> 2016-10-24 18:51:38 UTC https://padlet.com/kspangler/quadratic/wish/132825218 https://padlet.com/kspangler/quadratic/wish/132825436


]]> 2016-10-24 18:52:20 UTC https://padlet.com/kspangler/quadratic/wish/132825436 gradys1345 https://padlet.com/kspangler/quadratic/wish/132825604 x^2 + 4x + 20
the rad will be negative]]> 2016-10-24 18:52:52 UTC https://padlet.com/kspangler/quadratic/wish/132825604 https://padlet.com/kspangler/quadratic/wish/132825625 2016-10-24 18:52:55 UTC https://padlet.com/kspangler/quadratic/wish/132825625 https://padlet.com/kspangler/quadratic/wish/132825934 2016-10-24 18:53:43 UTC https://padlet.com/kspangler/quadratic/wish/132825934 maryamj2739 https://padlet.com/kspangler/quadratic/wish/132826127 x= -1 because (x+1)(x+1), the only solution to the problem is -1]]> 2016-10-24 18:54:19 UTC https://padlet.com/kspangler/quadratic/wish/132826127 alexisn1862 https://padlet.com/kspangler/quadratic/wish/132826144 the discriminant ends up being negative, making the solution an imaginary number]]> 2016-10-24 18:54:23 UTC https://padlet.com/kspangler/quadratic/wish/132826144 https://padlet.com/kspangler/quadratic/wish/132826333 2016-10-24 18:54:55 UTC https://padlet.com/kspangler/quadratic/wish/132826333 priscillab7507 https://padlet.com/kspangler/quadratic/wish/132826457 Trying to factor this equation by factoring would be impossible. Factoring requires you to have whole numbers or fractions as answers. However, if solved using the quadratic formula, you can find the real solutions.]]> 2016-10-24 18:55:20 UTC https://padlet.com/kspangler/quadratic/wish/132826457 https://padlet.com/kspangler/quadratic/wish/132826477 2016-10-24 18:55:24 UTC https://padlet.com/kspangler/quadratic/wish/132826477 https://padlet.com/kspangler/quadratic/wish/132826976 2016-10-24 18:57:04 UTC https://padlet.com/kspangler/quadratic/wish/132826976 briannar7340 https://padlet.com/kspangler/quadratic/wish/132827099 This equation factors out to (x-10)(x-2). By setting these equations equal to 0, you get the solutions 10 and 2, which are both real numbers, so there are two solutions.]]> 2016-10-24 18:57:28 UTC https://padlet.com/kspangler/quadratic/wish/132827099 https://padlet.com/kspangler/quadratic/wish/132827156 3x^2+6x+5=0
The quadratic equation can't be solved by factoring since 15 can't factor into 6. Only formula to use is the quadratic formula.]]> 2016-10-24 18:57:42 UTC https://padlet.com/kspangler/quadratic/wish/132827156 https://padlet.com/kspangler/quadratic/wish/132827172 2016-10-24 18:57:45 UTC https://padlet.com/kspangler/quadratic/wish/132827172 https://padlet.com/kspangler/quadratic/wish/132827566 2016-10-24 18:59:07 UTC https://padlet.com/kspangler/quadratic/wish/132827566 vanessal3761 https://padlet.com/kspangler/quadratic/wish/132827642 you can only solve this by using the quadratic formula b/c it cannot be factored. ]]> 2016-10-24 18:59:20 UTC https://padlet.com/kspangler/quadratic/wish/132827642 alexisn1862 https://padlet.com/kspangler/quadratic/wish/132827666 This equation has two real solutions that can be found by factoring because the 16 factors into the 10 evenly, ending up with (x+8) and (x+2). By setting both of the equations equal to zero, there are 2 real solutions. ]]> 2016-10-24 18:59:25 UTC https://padlet.com/kspangler/quadratic/wish/132827666 kevinl3942 https://padlet.com/kspangler/quadratic/wish/132827731 this equation does not factor because nothing multiplies to 8 and adds to 3. You can find the solutions using the quadratic formula]]> 2016-10-24 18:59:38 UTC https://padlet.com/kspangler/quadratic/wish/132827731 bradonw3792 https://padlet.com/kspangler/quadratic/wish/132827924 2016-10-24 19:00:14 UTC https://padlet.com/kspangler/quadratic/wish/132827924 imanim6925 https://padlet.com/kspangler/quadratic/wish/132828362 y=1
There is only one answer because when you get the two factors, they are the same. This is because there can only be so many factors for the number 2

]]> 2016-10-24 19:01:38 UTC https://padlet.com/kspangler/quadratic/wish/132828362 mariam3926 https://padlet.com/kspangler/quadratic/wish/132828403 x would equal ]]> 2016-10-24 19:01:47 UTC https://padlet.com/kspangler/quadratic/wish/132828403 nikolash2579 https://padlet.com/kspangler/quadratic/wish/132828567 2016-10-24 19:02:26 UTC https://padlet.com/kspangler/quadratic/wish/132828567 bobbyg6791 https://padlet.com/kspangler/quadratic/wish/132828911 if the numbers are plugged into the quadratic formula, then the the number that is inside the radical becomes a negative number. ]]> 2016-10-24 19:03:29 UTC https://padlet.com/kspangler/quadratic/wish/132828911 briannar7340 https://padlet.com/kspangler/quadratic/wish/132829003 This equation has two real solutions but cannot be factored into fractions or whole numbers. ]]> 2016-10-24 19:03:53 UTC https://padlet.com/kspangler/quadratic/wish/132829003 alexisn1862 https://padlet.com/kspangler/quadratic/wish/132829043 This equation has exactly one solution because it evenly factors into a perfect square, (x+3)^2. ]]> 2016-10-24 19:04:00 UTC https://padlet.com/kspangler/quadratic/wish/132829043 kevinl3942 https://padlet.com/kspangler/quadratic/wish/132829371 There is only one answer because when you factor the factors are both 3 .]]> 2016-10-24 19:04:53 UTC https://padlet.com/kspangler/quadratic/wish/132829371 vanessal3761 https://padlet.com/kspangler/quadratic/wish/132829452 when you solve this using the quadratic formula the discriminant is negative so the solution is an imaginary number]]> 2016-10-24 19:05:11 UTC https://padlet.com/kspangler/quadratic/wish/132829452 briannar7340 https://padlet.com/kspangler/quadratic/wish/132830171 The equation can't be factored but it has 2 real solutions ]]> 2016-10-24 19:07:28 UTC https://padlet.com/kspangler/quadratic/wish/132830171 tiffanyv4226 https://padlet.com/kspangler/quadratic/wish/132867158 This equation has one solution because b^2 would equal to 400 and subtracted by -4(10)(10) would result in a 0. The answer would then simplify to one solution. ]]> 2016-10-24 22:32:34 UTC https://padlet.com/kspangler/quadratic/wish/132867158 tiffanyv4226 https://padlet.com/kspangler/quadratic/wish/132867760 This equation would have two real solutions because once it is factored by getting (2x+4) or (x+2) and (2x+1) and making it equal to 0 there would be two solutions.]]> 2016-10-24 22:38:56 UTC https://padlet.com/kspangler/quadratic/wish/132867760 tiffanyv4226 https://padlet.com/kspangler/quadratic/wish/132868888 The equation could not be solved by factoring because there are no two numbers that would multiply to 18 and add to 8. This could be solved by using the quadratic formula. ]]> 2016-10-24 22:51:04 UTC https://padlet.com/kspangler/quadratic/wish/132868888 tiffanyv4226 https://padlet.com/kspangler/quadratic/wish/132869143 This equation would have 2 imaginary solutions because when solved by using the quadratic equation inside the radical the discriminant would turn out to be - number, which results in 2 imaginary solutions. ]]> 2016-10-24 22:53:47 UTC https://padlet.com/kspangler/quadratic/wish/132869143 priscillab7507 https://padlet.com/kspangler/quadratic/wish/132883029 If you subtract 16, you get the equation x^2=-16. When you try to square both sides, you would get x= radical -4. If you square a negative, it's imaginary because negatives cannot be squared]]> 2016-10-25 01:11:56 UTC https://padlet.com/kspangler/quadratic/wish/132883029 https://padlet.com/kspangler/quadratic/wish/133022347 2016-10-25 14:33:18 UTC https://padlet.com/kspangler/quadratic/wish/133022347 https://padlet.com/kspangler/quadratic/wish/133022574 x^2]]> 2016-10-25 14:33:49 UTC https://padlet.com/kspangler/quadratic/wish/133022574