Question 709676
Let x = the number
Then...
its square = {{{x^2}}}
twice its square = {{{2*x^2}}}
twice its square added to itself = {{{2*x^2 + x}}} (or {{{x + 2*x^2}}})<br>
So "twice its  square added to itself makes 10" translates into:
{{{2*x^2 + x = 10}}}
Now we solve this equation for x.<br>
This is a quadratic equation so we want one side to be zero. Subtracting 10 from each side:
{{{2*x^2 + x - 10 =0}}}
Next we factor (or use the Quadratic Formula). This factors without too much difficulty:
{{{(2x+5)(x-2) = 0}}}
From the Zero Product Property we know that this product can be zero <i>only</i> if one (or more) of the factors is zero. So:
2x+5 = 0 or x-2 = 0
Solving these we should get:
x = -5/2 or x = 2<br>
Since the problem asks for "whole number" solution we will reject -5/2. So the solution to your problem is 2.