Question

If x – y = 13 and xy = 28, then find x2 + y2.

Answer 1

(x-y)²=(13)²
(x)²+(y)²-2xy = 169
(x)²+(y)²=169+2xy
(x)²+(y)²=169+2×28
(x)²+(y)²=169+56
(x)²+(y)²=225

Answer 2

The value of $x^{2} +y^{2}=169.$

Step-by-step explanation:

We have,

x – y = 13 and xy = 28

To find, the value of $x^{2} +y^{2} =?$

We know tha,

$(x-y)^{2} =x^{2} +y^{2} -2xy$

⇒$x^{2} +y^{2}=(x-y)^{2}+2xy$     ....(1)

Put x – y = 13 and xy = 28 in (1), we get

$x^{2} +y^{2}=13^{2}+2(28)=169+56$

⇒ $x^{2} +y^{2}=225$

Hence, the value of $x^{2} +y^{2}=169.$