Jamierox4ev3r
  • Jamierox4ev3r
Find the composition of f(g(x)), when f(x)=7x^2-2x+9 and g(x)=x-4 Simplify
Mathematics
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Jamierox4ev3r
  • Jamierox4ev3r
Find the composition of f(g(x)), when f(x)=7x^2-2x+9 and g(x)=x-4 Simplify
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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Jamierox4ev3r
  • Jamierox4ev3r
I have a pretty clear idea of how to go about doing this, so I'll just show all of my process. I'll also point out the one small detail that I had a bit of doubt on.
Jamierox4ev3r
  • Jamierox4ev3r
\(7(x-4)^{2}-2(x-4)+9\) \(7(x-4)(x-4)-2x+8+9\) \((7x-28)(x-4)-2x+17\) \(7x^{2}-28x-28x+112-2x+17\) \(7x^{2}-58x+129\) ^^ I believe this is correct...but I forget the rules on difference of squares. does (x-2)^2 turn into (x-2)(x-2) or (x-2)(x+2) ?? Also, do I need to factor the final answer?
madhu.mukherjee.946
  • madhu.mukherjee.946
try and see if it can be done,then its good

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Jamierox4ev3r
  • Jamierox4ev3r
see it what can be done?
madhu.mukherjee.946
  • madhu.mukherjee.946
i don't think it can be factored
jim_thompson5910
  • jim_thompson5910
x^2 means x times x anything squared is that number multiplied by itself so (x-2)^2 means (x-2) times (x-2)
jim_thompson5910
  • jim_thompson5910
\[\Large 7x^{2}-58x+129\] is the correct final answer
Jamierox4ev3r
  • Jamierox4ev3r
Oh. You're referring to my question on whether it could be factor or not. oh and @jim_thompson5910 , so in the problem, if you look at my work, my assumption was indeed that (x-2)^2 was (x-2)(x-2). Thank you! I thought I remembered some funky rule from early high school about difference of squares or something like that. So I was correct all along. Wow I overthought this. Thank again!
jim_thompson5910
  • jim_thompson5910
you're thinking of a^2 - b^2 = (a-b)(a+b)
madhu.mukherjee.946
  • madhu.mukherjee.946
its perfectly all right
madhu.mukherjee.946
  • madhu.mukherjee.946
u can solve it by the formula
Jamierox4ev3r
  • Jamierox4ev3r
Oh okay. These rules are so easy to get confused, especially when trying to recall from memory
madhu.mukherjee.946
  • madhu.mukherjee.946
this formula is known as the SRIDHAR ACHARYA formula
madhu.mukherjee.946
  • madhu.mukherjee.946
don't worry by this formula it will get easy
madhu.mukherjee.946
  • madhu.mukherjee.946
put 7x^-58x+129 in this formula
Jamierox4ev3r
  • Jamierox4ev3r
Don't you mean the quadratic formula? And yes, I know that you can use the quadratic formula to factor. Problem is, I see now that when I use the quadratic formula, the roots are imaginary. Therefore, This cannot be factored any further. Jim's answer is the final answer
madhu.mukherjee.946
  • madhu.mukherjee.946
yeah
Jamierox4ev3r
  • Jamierox4ev3r
@madhu.mukherjee.946 thanks for trying to help, but I fully understand now. I've used the quadratic formula before, and I understand what it does \(\ddot\smile\)
madhu.mukherjee.946
  • madhu.mukherjee.946
okay.your welcome
jim_thompson5910
  • jim_thompson5910
You can use the discriminant formula as a shorter way to see if you can factor or not if D = b^2 - 4ac is a perfect square and a positive number, then you can factor
Jamierox4ev3r
  • Jamierox4ev3r
Woah, that's a helpful tip. *cough* I have a program in my calculator that solves all quadratics for me, so that's how I knew that the roots would be imaginary. But when in a pinch, when without a calculator, this is an extremely convenient tip. Thanks for bringing that back into my memory
jim_thompson5910
  • jim_thompson5910
definitely a handy thing to have
jim_thompson5910
  • jim_thompson5910
you're welcome

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