tonebeeze wrote:This PS quadratics problem on the OG Diagnostic really is a time sapper. Any suggestions regarding the most efficient way to solve these problem types?
If square root{3-2x} = square root{2x} +1, then 4x^2 =
(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1
*How do you add in mathematic symbols into our BTG posts?
OA = E
√(3-2x) = √2x + 1
[√(3-2x)]² = (√2x + 1)²
3 - 2x = (√2x + 1)(√2x + 1)
3 - 2x = 2x + √2x + √2x + 1
2 - 2x = 2x + 2√2x
1 - x = x + √2x
1 - 2x = √2x
(1 - 2x)² = (√2x)²
(1 - 2x)(1 - 2x) = 2x
1 - 2x - 2x + 4x² = 2x
4x² = 6x - 1.
The correct answer is
E.
To insert √ on a PC, hold down the Alt key as you type the numbers 251. Macs have a built-in character viewer that facilitates the insertion of math characters.
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