Find $25 x^{2}+16 y^{2}$, if $5 x+4 y=8$ and $x y=1$.
Given :
$5x+4y=8$ and $xy=1$.
To find :
We have to find the value of $25 x^{2}+16 y^{2}$.
Solution :
We know that, $(a+b)^2 = a^2+b^2+2ab$.
$(5x+4y)^2=25x^2 + 16y^2+2 (5x)(4y)$
$(5x+4y)^2= 25 x^2 + 16y^2+40xy$.........(i)
Substitute $5x+4y=8$ and $xy=1$ in (i),
$8^2 = 25x^2+16y^2+40(1)$
$64=25x^2+16y^2+40$
$64-40=25x^2+16y^2$
$25x^2+16y^2=24$
Therefore, the value of $25x^2+16y^2$ is 24.
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