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# What is the maximum value of x for which

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3019
What is the maximum value of x for which  [#permalink]

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31 Jul 2019, 05:14
00:00

Difficulty:

15% (low)

Question Stats:

88% (01:14) correct 12% (01:32) wrong based on 50 sessions

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Algerba Question Series- Question 2

What is the maximum value of x for which 8$$x^2$$ = 1 + 2x?

A. $$\frac{-1}{2}$$
B. $$\frac{-1}{4}$$
C. 0
D. $$\frac{1}{4}$$
E. $$\frac{1}{2}$$

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Joined: 19 Jun 2019
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Re: What is the maximum value of x for which  [#permalink]

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31 Jul 2019, 06:15
EgmatQuantExpert wrote:

Algerba Question Series- Question 2

What is the maximum value of x for which 8$$x^2$$ = 1 + 2x?

A. $$\frac{-1}{2}$$
B. $$\frac{-1}{4}$$
C. 0
D. $$\frac{1}{4}$$
E. $$\frac{1}{2}$$

\begin {alignat}{2} && 8x^2 = 1 + 2x \\ &\implies &8x^2 - 2x - 1 = 0 \\ &\implies &8x^2 - 4x + 2x -1 = 0 \\ &\implies &4x(2x - 1) + 1(2x - 1) = 0 \\ &\implies &(2x - 1)(4x + 1) = 0 \\ &\implies &x = \frac{1}{2} \text{ or } \frac{-1}{4} \end {alignat}

Maximum value of $$x$$ is $$\frac{1}{2}$$.
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Re: What is the maximum value of x for which  [#permalink]

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05 Aug 2019, 21:58

Solution

Given:
• An equation, $$8x^2 = 1 + 2x$$

To find:
• The maximum value of x for which $$8x^2 = 1 + 2x$$

Approach and Working Out:
$$8x^2 = 1 + 2x$$
• $$8x^2 - 2x – 1 = 0$$
• $$8x^2 – 4x + 2x – 1 = 0$$
• 4x * (2x – 1) + 1 * (2x – 1) = 0
• (4x + 1) * (2x – 1) = 0

Thus, x = $$\frac{1}{2} or –\frac{1}{4}$$

Therefore, the maximum value of x for which $$8x^2 = 1 + 2x$$ is $$\frac{1}{2}$$

Hence, the correct answer is Option E.

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Re: What is the maximum value of x for which   [#permalink] 05 Aug 2019, 21:58
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