Question

if x^2+y^2 = 40 and xy = 6 find the value of (x-y)^2​

Answer 1

Given: $x^2+y^2 = 40 \ and\ xy = 6$

We have to find the value of$(x-y)^2$.

As we know that $(a-b)^{2} =a^{2} +b^{2} -2ab$

We are solving in the following way:

We have,

$x^2+y^2 = 40 \ and\ xy = 6$

Applying the above identity, we are solving the above equation.

Now,

$\Rightarrow (x-y)^{2} =x^{2} +y^{2} -2xy\\\\\Rightarrow (x-y)^{2}=40-2\times6\\\\\Rightarrow (x-y)^{2}=40-12\\\Rightarrow (x-y)^{2}=28$

Hence$(x-y)^{2} =28$.