Finding the Positive Number A in a Parabola Problem

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alright so here is the problem.
The parabola y=x**2+3 has two tangents which pass through the point (0,–2). One is tangent to the to the parabola at (A,A2+3) and the other at (–A,A2+3). Find (the positive number) A.

i've attempted this problem several times but haven't been able to find the solution. :confused:
could someone tell me step by step how i would start this problem?
 
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Better to use x^2 rather than x**2 (I haven't seen that notation since BASIC). Any line through (0, -2) can be written y= mx- 2. A line tangent to y= x^2+ 3 at (A, A^2+ 3) must have slope m= 2A: y= 2Ax- 2
Determine A so that y= A^2+ 3= 2A(A)- 2= 3A^2- 2.
 
how did you get the slope to be 2A?
and what would the answer for A be?
 
The derivative of y= x^2 is y'= 2x. At x= A, y'= 2A and that is the slope ofthe tangent line.

Surely you can solve A^2+ 3= 3A^2- 2 yourself.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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