Question

 If (x +y) = 20 and xy = 12, then find the value ofx² + y²
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Answer 1

Answer:

(x+y)^2 = x^2 +y^2 +2xy

Given, x+y=20

xy=12 then

(20)^2 = x^2 + y^2 +2(12)

=> x^2 + y^2 =400 - 24

=> x^2 + y^2 = 376 Answer

Answer 2

Given: $(x +y) = 20 \ and\ xy = 12$

We have to find the value of$x^2+ y^2$.

By using the concept of identity rule that is$(a+b)^2=a^2+b^2+2ab$ we are solving the given problem.

We are solving in the following way:

We have,

$(x +y) = 20 \ and\ xy = 12$

From the identity rule, we can say that,

$(x+y)^2=x^2+y^2+2xy$

By putting the given values:

$=>(20)^2=x^2+y^2+2(12)\\=>400=x^2+y^2+24\\=>x^2+y^2=400-24\\=>x^2+y^2=376$

Hence, the value of $x^2+ y^2$will be$376$.