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Bunuel
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Hi Bunuel,

I understand how If g(x)=x^2 ... g(n+12) will thus be g(n+12)^2
However, I don't understand how if f(x)=2x-1 ... f(n^2) would be 2n^2-1 ... I thought it would be (2n-1)^2...why is this wrong?

Thanks.

Tosin
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ttaiwo
Hi Bunuel,

I understand how If g(x)=x^2 ... g(n+12) will thus be g(n+12)^2
However, I don't understand how if f(x)=2x-1 ... f(n^2) would be 2n^2-1 ... I thought it would be (2n-1)^2...why is this wrong?

Thanks.

Tosin

\(f(x) = 2x - 1\). To get \(f(n^2)\) put n^2 instead of x: \(f(n^2)=2n^2-1\).

\(g(x) = x^2\). To get \(g(n+12)\) put n + 12 instead of x: \(g(n+12)=(n+12)^2=n^2+24n+144\).
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I feel like I'm missing something important here. when I got to n^2-24n-145=0
I understand that the two values inside the (n-x)(n+y) will give us -145. but will the multiple values be the same if we find the values of n? how?
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I feel like I'm missing something important here. when I got to n^2-24n-145=0
I understand that the two values inside the (n-x)(n+y) will give us -145. but will the multiple values be the same if we find the values of n? how?

Solving n^2 - 24n - 145 = 0 gives n = -5 or n = 29 (n^2 - 24n - 145 = (n + 5)(n - 29)).
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I like the solution - it’s helpful.
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