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So I have a function: f(x)=nx^2+2x+n
And I have to put n like that so the function TOUCHES the x axis.
So I went on with this formula:
D=b^2-4ac
And if it touches that means that D=0.
0=4-4(n)(n)
0=4-4n^2
I am completely stumped here on what to do next.
4(1-n^2)=0
I tried this and the result is 1. But there was supposed to be 2 solutions to this problem ^.^. Any help greatly appriciated! Thank you.
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hi Smellyman,
What you have done is good.
So two things are multiplied and the product is zero => one of the things must be zero.
As it isn't the 4 it must be the other
There's the two solutions.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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How did I miss -1. Argh, silly me ^^. Thank you, kind sir! (:
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hi
You're welcome. Don't worry about missing that; it happens to us all at some time.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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