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What is the maximum value of x for which

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What is the maximum value of x for which  [#permalink]

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New post 31 Jul 2019, 05:14
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A
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C
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E

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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Re: What is the maximum value of x for which  [#permalink]

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New post 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 \(\frac{1}{2}\).
Answer is E.
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Re: What is the maximum value of x for which  [#permalink]

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New post 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.

Answer: 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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