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crimeas [40]
2 years ago
9

Let f(x)= x^2-4x-c. Find a nonzero value of c such that f(c)=c.​

Mathematics
1 answer:
Dimas [21]2 years ago
5 0

Answer:

The nonzero value of c will be:

  • c = 6

Step-by-step explanation:

Given the function

f\left(x\right)=\:x^2-4x-c

f\left(c\right)=\:c^2-4c-c

as

f(c) = c

so

c=\:c^2-4c-c

switching the sides

c^2-4c-c=c

subtract c from both sides

c^2-4c-c-c=c-c

c^2-6c=0

c\left(c-6\right)=0

Using the zero factor principle

\:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

c=0\quad \mathrm{or}\quad \:c-6=0

so, the solutions to the quadratic equations are:

c=0,\:c=6

Therefore, a nonzero value of c will be:

  • c = 6
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Step-by-step explanation:Simplifying

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Reorder the terms:

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Remove parenthesis around (-3 + 2j)

10 + -5j + -3 + 2j = 0

Reorder the terms:

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Combine like terms: 10 + -3 = 7

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Combine like terms: -5j + 2j = -3j

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Solving

7 + -3j = 0

Solving for variable 'j'.

Move all terms containing j to the left, all other terms to the right.

Add '-7' to each side of the equation.

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<h2><em><u>Hence , Proved</u></em>.</h2>

<h2><u>Hope it helps !</u></h2>

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