SOLUTION: find the whole number such that the twice its square added to itself makes 10

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Question 709676: find the whole number such that the twice its square added to itself makes 10

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the number
Then...
its square = x%5E2
twice its square = 2%2Ax%5E2
twice its square added to itself = 2%2Ax%5E2+%2B+x (or x+%2B+2%2Ax%5E2)

So "twice its square added to itself makes 10" translates into:
2%2Ax%5E2+%2B+x+=+10
Now we solve this equation for x.

This is a quadratic equation so we want one side to be zero. Subtracting 10 from each side:
2%2Ax%5E2+%2B+x+-+10+=0
Next we factor (or use the Quadratic Formula). This factors without too much difficulty:
%282x%2B5%29%28x-2%29+=+0
From the Zero Product Property we know that this product can be zero only 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

Since the problem asks for "whole number" solution we will reject -5/2. So the solution to your problem is 2.