Find the break even points of the expression $2x-3\left(x-3\right)^2$

Step-by-step Solution

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Final answer to the problem

$x=\frac{10+\sqrt{19}}{3},\:x=\frac{10-\sqrt{19}}{3}$
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Step-by-step Solution

How should I solve this problem?

  • Find break even points
  • Solve for x
  • Find the derivative using the definition
  • Solve by quadratic formula (general formula)
  • Simplify
  • Find the integral
  • Find the derivative
  • Factor
  • Factor by completing the square
  • Find the roots
  • Load more...
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Find the break even points of the polynomial $2x-3\left(x-3\right)^2$ by putting it in the form of an equation and then set it equal to zero

$2x-3\left(x-3\right)^2=0$

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$2x-3\left(x-3\right)^2=0$

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Learn how to solve problems step by step online. Find the break even points of the expression 2x-3(x-3)^2. Find the break even points of the polynomial 2x-3\left(x-3\right)^2 by putting it in the form of an equation and then set it equal to zero. A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words: (a-b)^2=a^2-2ab+b^2. Multiply the single term -3 by each term of the polynomial \left(x^2-6x+9\right). Combining like terms 2x and 18x.

Final answer to the problem

$x=\frac{10+\sqrt{19}}{3},\:x=\frac{10-\sqrt{19}}{3}$

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Function Plot

Plotting: $2x-3\left(x-3\right)^2$

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5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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