How do i solve 4(11-x) = (x-3)^2
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How do i solve 4(11-x) = (x-3)^2

[From: ] [author: ] [Date: 11-11-29] [Hit: ]
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Please find the solutions for this problem

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4(11-x) = (x-3)^2
4(11-x) = (x-3)(x-3)
4(11-x) = x^2 - 6x + 9
Then apply Distributive Property to left hand side of equation, so now you have:
44 - 4x = x^2 - 6x + 9
Add 4x to both sides:
44 = x^2 -2x + 9
Subtract 44 from both sides:
0 = x^2 - 2x - 35
Another way to put above equation is:
0 = (x + 5)(x - 7)

In order to find x, you set each "set" equal to zero and solve for x:
(x + 5) = 0 (x - 7) = 0
x = -5 x = 7


So the answer is x = -5, 7

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4(11 - x) --> 44 - 4x
(x - 3)^2 --> (x - 3)(x - 3) --> x^2 - 3x - 3x + 9 --> x^2 - 6x + 9

44 - 4x = x^2 - 6x + 9
(set equation equal to zero)

0 = x^2 - 2x - 35
(factor the trinomial)

(x - 7)(x + 5)

0 = -5, 7

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4(11-x) = (x-3)^2
x=7,-5

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4(11 - x) = (x - 3)^2
44 - 4x = x^2 - 6x + 9
x^2 - 2x - 35 = 0
(x - 7)(x + 5) = 0
x = -5,7
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