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,

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

$\Rightarrow x^{2} + 81 - 18 x + x^{2} = 261$ $rArr x^2+81-18x+x^2 = 261$

$\Rightarrow 2 x^{2} - 18 x - 180 = 0$ $rArr 2x^2-18x-180 = 0$

$\Rightarrow x^{2} - 9 x - 90 = 0$ $rArr x^2-9x-90 = 0$

$\Rightarrow x^{2} - 15 x + 6 x - 90 = 0$ $rArr x^2-15x+6x-90 = 0$

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

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

$\Rightarrow x = 15 or - 6$ $rArr x = 15 or -6$

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

Answer 2 · 2017-04-28T16:31:11

$- 6 and 15$ $-6 " and " 15$

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

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

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

$x = 9 - y$ $x = 9 - y$

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

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

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

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

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

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

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

Solve for $y$ $y$ :

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

Solve for $x$ $x$ :

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

Solution: $- 6 and 15$ $-6 " and " 15$