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If $x + y = 4$ and $xy = 2$, find the value of $x^2 + y^2$.
Given:
$x + y = 4$ and $xy = 2$
To do:
We have to find the value of $x^2 + y^2$.
Solution:
The given expressions are $x + y = 4$ and $xy = 2$. Here, we have to find the value of $x^2 + y^2$. So, by squaring the given expression and using the identity $(a+b)^2=a^2+2ab+b^2$, we can find the required value.
$xy = 2$...........(i)
$(a+b)^2=a^2+2ab+b^2$...........(ii)
Now,
$x + y = 4$
Squaring on both sides, we get,
$(x + y)^2 = 4^2$
$x^2+2 \times x \times y+y^2=16$ [Using (ii)]
$x^2+2xy+y^2=16$
$x^2+2(2)+y^2=16$ [Using (i)]
$x^2+4+y^2=16$
$x^2+y^2=16-4$ (Transposing 4 to RHS)
$x^2+y^2=12$
Hence, the value of $x^2+y^2$ is $12$.
Updated on: 01-Apr-2023
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