find x^2+y^2 (1 Viewer)

[_kuroo]

need help,

if x+y=1 and x^3+y^3=19 , how do i find the value of x^2+y^2?

 

[deswa1]

For this question, the easiest way to do it would be to solve both of those equations simultaneously as the numbers work out nicely giving you (3,-2) as the values for x and y (or vice versa). Then simply square both of those and your done.

 

[_kuroo]

umms yeah.....i still dunno how LOLS

could you write the steps down?

 

[Jason Xie]

because: x^3+y^3=19
therefore : (x+y)(x^2+xy+y^2)=19 (factorise x^3+y^3=19)
because x+y=1 (given)
therefore: x^2+xy+y^2 = 19 this lead to x^2 + y^2 = 19 - xy...................(1)
because: x^2 + 2xy + y^2 =1 , this lead to x^2 + y^2 = 1 - 2xy...............(2)
equation (1) - equation (2) : 18 + xy = 0
therefore we know xy=-18
substitude xy=-18 into (1)
x^2 + y^2 = 19 - (-18)
= 37

This is my method, if u want to solve simultaneously, the working process is not as good as this

 

[Jason Xie]

sorry, i did something wrong!! (x+y)(x^2+xy+y^2)=19 (factorise x^3+y^3=19)
should be (x+y)(x^2-xy+y^2)=19
therefore: x^2-xy+y^2 = 19 this lead to x^2 + y^2 = 19 + xy...................(1)
because: x^2 + 2xy + y^2 =1 , this lead to x^2 + y^2 = 1 - 2xy...............(2)
equation (1) - equation (2) : 18 + 3xy = 0
therefore we know xy=-6
substitude xy=-6 into (1)
x^2 + y^2 = 19 -6= 13

 

[RealiseNothing]

A more intuitive way would be to realise that we could make x a positive number and y a negative, and in doing so their cubes must differ by 19.

It's obvious now what the values for x and y are: x=3 and y=-2 , as the difference of their cubes is 27 - 8 = 19.

Hence x^2 + y^2 = 9 + 4 = 13

 

[deswa1]

umms yeah.....i still dunno how LOLS

could you write the steps down?

This is how I did it.

<a href="http://www.codecogs.com/eqnedit.php?latex=1-y=x\\ 1-3y@plus;3y^2-y^3@plus;y^3=19 \\3y^2-3y-18=0 \\3(y-3)(y@plus;2)=0\\ y=-2,3\\ \therefore x^2@plus;y^2=(-2)^2@plus;3^3=13" target="_blank"><img src="http://latex.codecogs.com/gif.latex?1-y=x\\ 1-3y+3y^2-y^3+y^3=19 \\3y^2-3y-18=0 \\3(y-3)(y+2)=0\\ y=-2,3\\ \therefore x^2+y^2=(-2)^2+3^3=13" title="1-y=x\\ 1-3y+3y^2-y^3+y^3=19 \\3y^2-3y-18=0 \\3(y-3)(y+2)=0\\ y=-2,3\\ \therefore x^2+y^2=(-2)^2+3^3=13" /></a>

 

[_kuroo]

thanks soo much guys

 

[_kuroo]

i got a test tomorrow, so i might be asking a few more questions on the forum about stuff im not sure about. i have this lousy teacher that skips alot of steps inbetween stuff when his explaining and just writes the question and then answer.

 

[deswa1]

i got a test tomorrow, so i might be asking a few more questions on the forum about stuff im not sure about. i have this lousy teacher that skips alot of steps inbetween stuff when his explaining and just writes the question and then answer.

Yeah go for it.