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3 years ago

If (x-y)^2=100 and xy=20. what's the value of x^2+y^2?

2 answers:

8 0

$(x-y)^2=x^2-2xy+y^2=(x^2+y^2)-2(xy)\\\\\text{We have}\ (x-y)^2=100\ \text{and}\ xy=20.\\\\\text{Substitute}\\\\100=(x^2+y^2)-2(20)\\100=(x^2+y^2)-40\qquad\text{add 40 to both sides}\\\\140=(x^2+y^2)\\\\Answer:\ \boxed{x^2+y^2=140}$

Flauer [41]3 years ago

7 0

Answer: The value of $x^2+y^2$ is 140

Step-by-step explanation:

We are given an expression:

$(x-y)^2=100$

And,

xy = 20

Using the identity:

$(a-b)^2=a^2+b^2-2ab$

Solving the expression:

$x^2+y^2-2xy=100$

Putting the value of 'xy' in above equation, we get:

$x^2+y^2-2(20)=100\\\\x^2+y^2=100+40\\\\x^2+y^2=140$

Hence, the value of $x^2+y^2$ is 140

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by
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+
10
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4
+
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3
+
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4
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x
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⋅
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x
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⋅
2
3
+
5
x
⋅
2
4
+
2
5
Multiply
4
4
by
10
10
.
x
5
+
10
x
4
+
40
x
3
+
10
x
2
⋅
2
3
+
5
x
⋅
2
4
+
2
5
x
5
+
10
x
4
+
40
x
3
+
10
x
2
⋅
2
3
+
5
x
⋅
2
4
+
2
5
Raise
2
2
to the power of
3
3
.
x
5
+
10
x
4
+
40
x
3
+
10
x
2
⋅
8
+
5
x
⋅
2
4
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x
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8
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by
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3
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⋅
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⋅
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4
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x
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40
x
3
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80
x
2
+
80
x
+
2
5
x
5
+
10
x
4
+
40
x
3
+
80
x
2
+
80
x
+
2
5
Raise
2
2
to the power of
5
5
.
x
5
+
10
x
4
+
40
x
3
+
80
x
2
+
80
x
+
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