The of two numbers is 9 and their product is 20- Find the of their squares-displaystyle 39displaystyle 41displaystyle 26displaystyle 13

Question

Toppr Admin · Accepted Answer

Let two number be x and yGiven-x-y-9 andxy-20Now-x-y-9Squaring both the sides-x-y-2-92-gt-x2-y2-2xy-81-gt-x2-y2-2-20-81-gt-x2-y2-40-81-gt-x2-y2-81-x2212-40-gt-x2-y2-41

The sum of two numbers is 9 and their product is 20. Find the sum of their squares.
$39$
$41$
$26$
$13$

Let two number be $x$ and $y$
Given,
$x + y = 9$ and
$x y = 20$
Now,
$x + y = 9$
Squaring both the sides,
$(x + y)^{2} = 9^{2}$
$=> x^{2} + y^{2} + 2 x y = 81$
$=> x^{2} + y^{2} + 2 (20) = 81$
$=> x^{2} + y^{2} + 40 = 81$
$=> x^{2} + y^{2} = 81 - 40$
$=> x^{2} + y^{2} = 41$

The sum of two numbers is 9 and their product is 20. Find the sum of their squares.39412613

Let two number be x and yGiven,x+y=9 andxy=20Now,x+y=9Squaring both the sides,(x+y)2=92=>x2+y2+2xy=81=>x2+y2+2(20)=81=>x2+y2+40=81=>x2+y2=81−40=>x2+y2=41

The sum of two numbers is 9 and their product is 20. Find the sum of their squares.
$39$
$41$
$26$
$13$

Let two number be $x$ and $y$
Given,
$x + y = 9$ and
$x y = 20$
Now,
$x + y = 9$
Squaring both the sides,
$(x + y)^{2} = 9^{2}$
$=> x^{2} + y^{2} + 2 x y = 81$
$=> x^{2} + y^{2} + 2 (20) = 81$
$=> x^{2} + y^{2} + 40 = 81$
$=> x^{2} + y^{2} = 81 - 40$
$=> x^{2} + y^{2} = 41$