Question

The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers.​

Answer 1

Step-by-step explanation:

let the numbers be x & y

x+y = 22

x= 22-y

x^2+y^2=250

..

plug value of x in the equation

(22-y)^2+y^2=250

484-44y+y^2+y^2=250

2y^2-44y+484-250=0

2y^2-44y+234=0

/2

y^2-22y+117=0

x^2-13y-9y+117=0

x(x-13)-9(x-13)=0

(x-13)(x-9)=0

x= 13 OR 9

the numbers are 9 & 13

Answer 2

Step-by-step explanation:

Given: The sum of two numbers is 22 and the sum of their squares is 250.

To find: Find the numbers.

Solution:

Let the numbers are x and y.

Step 1: Write equations according to the given conditions.

$\bf \red{x + y = 22...eq1}$

$\bf \green{{x}^{2} + {y}^{2} = 250 \: ...eq2}$

Step 2: Put the value of y from eq1 into eq2

$y = 22 - x$

${x}^{2} + ( {22 - x)}^{2} = 250$

open identity

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

So,

${x}^{2} + 484 - 44x + {x}^{2} = 250$

or

$2 {x}^{2} - 44x + 484 - 250 = 0$

$2 {x}^{2} - 44x + 234 = 0$

cancel common term

${x}^{2} - 22x + 117 = 0$

${x}^{2} - 13x -9x+ 117 =0$

${x}(x - 13) -9(x-13)=0$

$(x-9)(x - 13) =0$

$(x-9) =0$

$\bf x=9$

or

$(x-13) =0$

$\bf x=13$

Step 3: Find the numbers.

Case 1: When $x=9$

$9 + y = 22$

$y = 13$

Case 2: When $x=13$

$13 + y = 22$

$y = 9$

Verification:

$9 + 13= 22$

$(9)^2 + (13)^2= 250$

$81 + 169= 250$

Final answer:

The numbers are 9 and 13.

Hope it helps you.

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