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Bunuel
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Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of al 28 Jun 2022, 01:30
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Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of all values of z for which f(3z) = 7.


(A) -1/3
(B) -1/9
(C) 0
(D) 5/9
(E) 5/3



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B
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Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of al 28 Jun 2022, 01:55
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f(3z)=f((9z)/3)=(9z)^2+9z+1=7
=> 81z^2 + 9z -6 =0
=>(9z+3)(9z-2)=0
So either 9z+3=0 Or 9z-2=0
=> z=-3/9 Or z=2/9
Sum of -3/9 and 2/9 is -1/9

The answer is thus (B).

Note: at 81z^2 + 9z -6 =0
we can also go directly to calculate z1+z2=-(9/81)=-1/9
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Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of al 26 Jul 2022, 08:15
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Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of all values of z for which f(3z) = 7.


(A) -1/3
(B) -1/9
(C) 0
(D) 5/9
(E) 5/3


\(f(\frac{9z}{3})=f(3z)\), so we can plug 9z in for x.

\((9z)^2+9z+1 = 7\)

\(81z^2+9z+1 = 7\)

\(81z^2+9z-6 = 0\)

\((9z+3)(9z-2) = 0\)

\(9z+3 = 0\) ... OR ... \(9z-2 = 0\)

\(9z = -3\) ... OR ... \(9z = 2\)

\(z = \frac{-3}{9}\) ... OR ... \(z = \frac{2}{9}\)

\(\frac{-3}{9}+\frac{2}{9} = \frac{-1}{9}\)

Answer choice B.
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Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of al 07 Dec 2024, 05:49
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧




f(x/3) = \(x^2 + x + 1\) and we need to find all values of z for which f(3z) = 7

To find f(3z) we need to compare what is inside ( ) in f(x/3) = \(x^2 + x + 1\) and f(3z)

=> We need to substitute x/3 with 3z in f(x/3) = \(x^2 + x + 1\) to get the value of f(3z)

x/3 = 3z
=> x = 3*3z = 9z

=> f(9z/3) = f(3z) = \((9z)^2 + 9z + 1\) = 7
=> 81\(z^2\) + 9z - 6 = 0
=> 27\(z^2\) + 3z - 2 = 0
=> 27\(z^2\) + 9z - 6z - 2 = 0
=> 9z*(3z + 1) - 2(3z + 1) = 0
=> (9z - 2) * (3z + 1) = 0
=> z = \(\frac{2}{9}\), \(\frac{-1}{3}\)

=> Sum of all values of z = \(\frac{2}{9}\) + \(\frac{-1}{3}\) = \(\frac{2}{9}\) + \(\frac{-3}{9}\) = \(\frac{-1}{9}\)

So, Answer will be B
Hope it helps!

Watch the following video to MASTER Functions and Custom Characters

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Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of al 08 Dec 2024, 09:20
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Bunuel
Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of all values of z for which f(3z) = 7.


(A) -1/3
(B) -1/9
(C) 0
(D) 5/9
(E) 5/3



Are You Up For the Challenge: 700 Level Questions­
I spent 10 mins while finding 3 possible ways to solve the possible values of z once we get the quadratic eqn 81z^2 + 9z - 6 = 0
1. Solve the eqn and we will get two roots -(1/3) and (2/9).
2. Do not solve the eqn, substitute values lol.
3. Since the question asks to find the sum of the possible values z, we can tell we will have only two roots since this is a quad eqn. There is no need to find any roots as well. We know, the sum of roots of a quad eqn is given by -(b/a) where eqn is ax^2 + bx + c =0, simply put b = 9 and a = 81 we will get -(1/9) which is option (B).
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Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of al 09 Dec 2024, 11:33
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it has a trick i found out. FOr any sum of all solutions of a given quadratic equation like the above question, it has to be -b/a
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Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of al 09 Dec 2024, 20:10
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Bunuel
Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of all values of z for which f(3z) = 7.


(A) -1/3
(B) -1/9
(C) 0
(D) 5/9
(E) 5/3



Are You Up For the Challenge: 700 Level Questions­
The solutions in the comments are unnecessarily magnifying the quadratic equation and then solving it and obtaining the roots. You could instead let z = x/9, then f(3z) = f(x/3) = x^2 + x+ 1 = 7

So the quadratic is x^2 +x - 6 = 0, which is much more easily factorised as (x+3)(x-2). Then x is -3 and 2. If z = x/9, then z = -3/9 and 2/9 which when added gives -1/9.
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Let f be a function for which f(x/3) = x^2 + x + 1. Find the sum of al 09 Dec 2024, 20:10
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mini97
it has a trick i found out. FOr any sum of all solutions of a given quadratic equation like the above question, it has to be -b/a
Its Viete's formula :)
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