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2 solutions
thanks
i didnt get it can u explain it again in detail?
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It is given in the problem that,
x
+
y
=
1
x 3 + y 3 = 2 5
Now, there is a well known expansion/formula for the second equation, namely,
x
3
+
y
3
=
(
x
+
y
)
(
x
2
+
y
2
−
x
y
)
You can easily confirm that it is true by just multiplying out the brackets.
So,
x
3
+
y
3
=
(
x
+
y
)
(
x
2
+
y
2
−
x
y
)
=
2
5
But, it is given that x + y = 1 , so,
( x + y ) ( x 2 + y 2 − x y ) = 2 5
( 1 ) ( x 2 + y 2 − x y ) = 2 5
x 2 + y 2 − x y = 2 5 − − − − ( 1 )
Now,
x + y = 1
⇒ ( x + y ) 2 = 1 2
⇒ x 2 + y 2 + 2 x y = 1 − − − − ( 2 )
Solving ( 1 ) and ( 2 ) for x y and x 2 + y 2 (I hope you know how to solve two equations, if not, just ask and i'll explain it to you),
x 2 + y 2 = 1 7 x y = − 8
Finally,
We need to find out the value of
x
4
+
y
4
...To do this, we just do the following...
x 4 + y 4 = ( x 2 ) 2 + ( y 2 ) 2 = ( x 2 + y 2 ) 2 − 2 x 2 y 2 (If you do not understand the above part, please just ask and i'll explain it to you)
Now, we just substitute the values of
x
2
+
y
2
and (xy) that we just found out,
⇒
x
4
+
y
4
=
(
1
7
)
2
−
2
(
−
8
)
2
=
2
8
9
−
1
2
8
=
1
6
1
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im sorry.. but do you mind explaining the solving for the 1 &2 equation or the solving of two equation.. pl z
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@John Warren Mejico
–
Referring to my comment above,
x
2
+
y
2
−
x
y
=
2
5
−
−
−
−
(
1
)
x
2
+
y
2
+
2
x
y
=
1
−
−
−
−
−
(
2
)
Performing
(
2
)
−
(
1
)
,
3
x
y
=
−
2
4
⇒
x
y
=
−
8
Now, just substitute this value in any one of the equations...Its your choice..either ( 1 ) or ( 2 )
I will substitute the value in ( 1 ) (please remember, you can choose any one of the equations..)
Substituting in
(
1
)
,
x
2
+
y
2
−
(
−
8
)
=
2
5
x
2
+
y
2
+
8
=
2
5
x
2
+
y
2
=
1
7
Hence, we get the 2 values which I wrote in my previous comment.
excellant
that is good.....
What is the value of x and y.
x+y=1, x3+y3=25 or (x+y)(x2-xy+y2)=25 or (x+y)((x+y)2-3xy)=25 or 1((1-3xy)=25 or xy=-8 now x4+y4=(x2)2+(y2)2=(x2+y2)2-2x2y2=((x+y)2-2xy)2-2(-8)2=(1+16)2-128=289-128=161
x + y = 1 x 3 + y 3 = 2 5
⇒ ( x + y ) ( x 2 + y 2 − x y ) = 2 5 ⇒ x 2 + y 2 − x y = 2 5 − − − − − ( 1 ) Also,
x 2 + y 2 + 2 x y = 1 − − − − − − ( 2 )
Using ( 1 ) and ( 2 ) ,
⇒ x y = − 8 ⇒ x 2 + y 2 = 1 7
Now,
x 4 + y 4 = ( x 2 + y 2 ) 2 − 2 x 2 y 2 = ( 1 7 ) 2 − 2 ( − 8 ) 2 = 1 6 1