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Another Algebra 2 question...

 Oct 3, 2017
 #1
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Solve for x:
(3 x - 2)^2 = (5 x + 1)^2

Take the square root of both sides:
3 x - 2 = 5 x + 1 or 3 x - 2 = -5 x - 1

Subtract 5 x - 2 from both sides:
-2 x = 3 or 3 x - 2 = -5 x - 1

Divide both sides by -2:
x = -3/2 or 3 x - 2 = -5 x - 1

Add 5 x + 2 to both sides:
x = -3/2 or 8 x = 1

Divide both sides by 8:
x = -3/2         or          x = 1/8

 Oct 3, 2017
 #2
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"Use the identity a2-b2=(a-b)(a+b) to solve the problem:  (3x-2)2=(5x+1)2"

 

Rewrite as:  (5x+1)2 - (3x-2)2 = 0

 

Let a = 5x+1 and b = 3x-2

 

(5x+1)2 - (3x-2)2 → (5x+1 - (3x-2))(5x+1 + 3x-2) → (2x + 3)(8x - 1)

 

so (2x + 3)(8x - 1) = 0

 

Hence 2x + 3 = 0   giving  x = -3/2

 

or  8x - 1 = 0  giving  x = 1/8

.

 Oct 3, 2017

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