Question

The sum of a number and its square is 182. Find the number.

Answer 1

Answer:

13 and -14 satisfy this condition

Step-by-step explanation:

Let's represent that number as x

and the square of x is x^2

So,

x + x^2 = 182

Subtract 182 from both sides

x + x^2 - 182 = 182 - 182

x + x^2 - 182 = 0

rearrange the quadratic equation

x^2 + x -182 = 0

let's use the quadratic formula

$\frac{-b+\sqrt{b^2-4ac} }{2a}$       or      $\frac{-b-\sqrt{b^2-4ac} }{2a}$

a = 1, b = 1, c = -182

$\frac{-1+\sqrt{1^2-4*1*(-182)} }{2*1}$       or     $\frac{-1-\sqrt{1^2-4*1*(-182)} }{2*1}$

$\frac{-1+\sqrt{1+728} }{2}$      or         $\frac{-1-\sqrt{1+728} }{2}$

$\frac{-1+\sqrt{729} }{2}$            or    $\frac{-1-\sqrt{729} }{2}$

$\frac{-1+{27} }{2}$     or    $\frac{-1-{27} }{2}$

$\frac{26}{2}$     or    $\frac{-28}{2}$

13 or    - 14

Lets check

13 + 13^2 = 13 + 169

= 182

Also,

-14 + (-14^2) = -14 + 196

= 182