Given #x+y=2# and #x^3+y^3=5#, what is #x^2+y^2#?

Question

Eva Adamson · Accepted Answer

It is #x^2+y^2=3#

From #x+y=2# we have that

#x+y=2=>(x+y)^2=2^2=>x^2+y^2+2xy=4#

From #x^3+y^3=5# we have that

#x^3+y^3=5=>(x+y)*(x^2+y^2-xy)=5=> x^2+y^2-xy=5/2#

Thus, we are aware that

#x^2+y^2+2xy=4# (1)

additionally

#x^2+y^2-xy=5/2=>2*(x^2+y^2)-2xy=5# (2)

Adding (1) and (2) results in

#3*(x^2+y^2)=9=>x^2+y^2=3#

Abigail Allen · Answer

#x^2 + y^2 = 3#

By combining the provided equations with the sum of cubes formula:

#5 = x^3 + y^3 = (x+y)(x^2-xy+y^2) = 2(x^2 + y^2 - xy)#

#=> x^2 + y^2 - xy= 5/2#

#=>x^2 + y^2 = 5/2 + xy" "("*")#

Now, if we cube the first equation provided, we obtain

#(x+y)^3 = 2^3#

#=>x^3 + 3x^2y + 3xy^2 + y^3 = 8#

#=> (x^3 + y^3) + 3xy(x + y) = 8#

Substituting in our known values for #x+y# and #x^3 + y^3#, we obtain

#=> 5 + 3xy*2 = 8#

#=> 6xy = 3#

#=> xy = 1/2#

Substituting this back into #("*")#:

#x^2 + y^2 = 5/2 + 1/2#

#:. x^2 + y^2 = 3#

Isaiah Anthony · Answer

undefined To find $x^2 + y^2$, we can use the following algebraic manipulations:

Given $x + y = 2$, we can square this equation to get $x^2 + 2xy + y^2 = 4$.

Next, we can substitute $x^3 + y^3 = 5$ into the equation $x^2 + 2xy + y^2 = 4$.

This substitution gives us:

$x^2 + 2xy + y^2 = (x + y)(x^2 - xy + y^2) = 4$.

Since $x + y = 2$, we have:

$2(x^2 - xy + y^2) = 4$.

Dividing by 2, we get:

$x^2 - xy + y^2 = 2$.

Multiplying $x + y = 2$ by $x^2 - xy + y^2 = 2$, we get:

$(x + y)(x^2 - xy + y^2) = 2 \times 2$.

Substituting the values we have:

$2 \times 2 = 4$.

Thus, $x^2 + y^2 = 4$.