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

Apr 28, 2017

Either 15 and -6 or -6 and 15.

#### Explanation:

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

As per questions,

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

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

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

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

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

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

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

$\Rightarrow x = 15 \mathmr{and} - 6$

other's number is $\left(9 - 15\right) = - 6 \mathmr{and} \left(9 + 6\right) = 15$
Hence numbers are either 15 and -6 or -6 and 15.

Apr 28, 2017

$- 6 \text{ and } 15$

#### Explanation:

Given: Sum of the two numbers is $9 : \text{ } x + y = 9$

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

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

$x = 9 - y$

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

Distribute using ${\left(a - b\right)}^{2} = \left({a}^{2} - 2 a b + {b}^{2}\right)$:

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

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

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

Factor a $2 : \text{ } 2 \left({y}^{2} - 9 y + 90\right) = 0$

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

Solve for $y$:

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

Solve for $x$:

$\text{ "x = 9 - 15 = -6 " and } x = 9 - - 6 = 9 + 6 = 15$

Solution: $- 6 \text{ and } 15$