The sum of two numbers is 9 and the sum of their squares is 261. How do you find the numbers?

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Answer 1 · 2017-04-28T16:26:30

Either 15 and -6 or -6 and 15.

Let one number be x, then other's is 9 - x.

As per questions,

$x^2+(9-x)^2 = 261$

$rArr x^2+81-18x+x^2 = 261$

$rArr 2x^2-18x-180 = 0$

$rArr x^2-9x-90 = 0$

$rArr x^2-15x+6x-90 = 0$

$rArr x(x-15)+6(x-15) = 0$

$rArr (x-15)(x+6)= 0$

$rArr x = 15 or -6$

other's number is $(9 - 15) = - 6 or (9+ 6) = 15$
Hence numbers are either 15 and -6 or -6 and 15.

Answer 2 · 2017-04-28T16:31:11

$-6 " and " 15$

Given: Sum of the two numbers is $9: " "x + y = 9$

Given: Sum of the squares is $261: " "x^2 + y^2 = 261$

Use substitution by solving for one variable in the first equation and substituting this variable into the second equation:

$x = 9 - y$

$(9 - y)^2 + y^2 = 261$

Distribute using $(a - b)^2 = (a^2 - 2ab + b^2)$ :

$81 - 18y + y^2 + y^2 = 261$

Simplify: $2y^2 - 18y +81 - 261 = 0$

$2y^2 - 18y +180 = 0$

Factor a $2: " "2(y^2 - 9y + 90) = 0$

Factor quadratic: $2(y - 15)(y + 6) = 0$

Solve for $y$ :

$y - 15 = 0; " " y = 15 " and " y + 6 = 0; " " y = -6$

Solve for $x$ :

$" "x = 9 - 15 = -6 " and " x = 9 - -6 = 9 + 6 = 15$

Solution: $-6 " and " 15$