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31 Jul 2019, 05:14
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
79% (01:28) correct
21%
(01:43)
wrong
based on 188
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. \(\frac{1}{2}\)
31 Jul 2019, 06:15
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EgmatQuantExpert
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.
05 Aug 2019, 21:58
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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
