This isn't a homework service. You have to show that you at least made an effort to try them rather than just getting other people to do it for you.
We need proof that you actually tried, and then we can help you on parts that you get stuck on or cannot do.
i can do for number 1 only
x^4 - 11x^2 + 28 = 0
(x^2 -7 ) ( x^2 -4 ) =
(x^2 - 7 ) 0 | (x^2 - 4) = 0
x^2 = 7 | x^2 = 4
x = = + - sq. root of 7| x = +- of 2
i dnt knw the rest beacuse i only knw the simple special products and factoring patterns.
2. x^-4 - 13x^-2 + 36 = 0
( x^-2 + 6 )(x^-2 + 6 ) =
i dnt knw next because it has a negative exponents and some of the rest of the given... sorry
rotchkaroots lang.
Go back to x^-4- 13x^-2+ 36= 0.
There are two ways to handle that:
1) Let y= x^-2. Then x^-4= y^2 so the equation becomes y^2- 13y+ 36= 0. Once you have found y, x= 1/sqrt(y).
2) Multiply the entire equation by x^4. You get 1- 13x^2+ 36x^4= 0 which becomes a quadratic by letting y= x^2: 36y^2- 13y+ 1= 0. Once you have found y, x= sqrt(y).
i`m sorry if i have made u think that way (Educated) i just needed help. i never needed an insult sir/madam. i also tried my best to solve the problems in my own little way. I thank you so much for your help. it just happend that i wasn't able to show to you how i did it. Sorry.
You can check them yourself just by putting the numbers into the original equation. If x= -23/9, then x+ 3= -23/9+ 27/9= 4/9 and .
so that is correct.
If x= -2, then x+ 3= 1 and .
, not 2, so that is not correct.
The reason you got an incorrect answer was that, after getting M= 2/3 and -1 as the (correct) solutions to the quadratic, you set which is impossible- a square root is never negative.