# Prove that this quadratic is greater than 0Watch

#1
I don't understand the alternative method shown in the pic below.

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#2
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1 week ago
#3
(Original post by JacobBob)
I don't understand the alternative method shown in the pic below.

A parabola with positive x^2 coefficient (i.e. ax^2 where a>0. In this case a=1) is "U" shaped. So, if the minimum value is greater than 0, so is the rest of it.

Min value is at x=-1/2, and it's greater than zero, hence....
1 week ago
#4
The lowest point of a parabola is the vertex/ min point. To find it dy/dx =0 and at this point x^2+x+1 s positive, if the minimum is positive and every otger value on the parabola is greater than the minimum, then all points are positive.
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1 week ago
#5
They differentiated the equation to find the gradient at a given point
They set this equation to 0, so they found the x coordinate of the minimum point (one of the turning points) of the graph
They then put this x coordinate back into the original equation to find the y coordinate of the minimum point
This was greater than 0, so for all values of x, the quadratic is greater than 0.
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