Given -x - y- - 13 and xy - 22- the value of -x-2- - -y-2-

Question

Toppr Admin · Accepted Answer

X-y-13xy-22Squaring bothside weget-xA0-x-y-2-132x2-y2-2xy-169-Putting the value of xy in-xA0-x2-y2-2-xD7-22-169x2-y2-144-x2212-44x2-y2-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-xA0-

Given $(x + y) = 13$ and $x y = 22$ , find the value of $x^{2} + y^{2}$

$x + y = 13$
$x y = 22$

Squaring bothside weget
.
$(x + y)^{2} = 13^{2} x^{2} + y^{2} + 2 x y = 169.$ .

Putting the value of $x y$ in

$x^{2} + y^{2} + 2 \times 22 = 169$

$x^{2} + y^{2} = 144 - 44 x^{2} + y^{2}$

Given (x+y)=13 and xy=22, find the value of x2+y2

x+y=13xy=22Squaring bothside weget .(x+y)2=132x2+y2+2xy=169..Putting the value of xy in x2+y2+2×22=169x2+y2=144−44x2+y2

Given $(x + y) = 13$ and $x y = 22$ , find the value of $x^{2} + y^{2}$

$x + y = 13$
$x y = 22$

Squaring bothside weget
.
$(x + y)^{2} = 13^{2} x^{2} + y^{2} + 2 x y = 169.$ .

Putting the value of $x y$ in

$x^{2} + y^{2} + 2 \times 22 = 169$

$x^{2} + y^{2} = 144 - 44 x^{2} + y^{2}$