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If z ≠ 0 and \(z + \frac{1 - 2z^2}{z} =\frac{ w}{z}\), then\( w = \)
A. \(z + 1 \)
B. \(z^2 + 1\)
C. \(-z^2 + 1\)
D. \(-z^2 + z + 1\)
E. \(-2z^2 + 1\)
PS14031.02
A. \(z + 1 \)
B. \(z^2 + 1\)
C. \(-z^2 + 1\)
D. \(-z^2 + z + 1\)
E. \(-2z^2 + 1\)
PS14031.02
20 Apr 2020, 11:28
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gmatt1476
If z ≠ 0 and \(z + \frac{1 - 2z^2}{z} = w/z\), then w =
A. z + 1
B. z^2 + 1
C. -z^2 + 1
D. -z^2 + z + 1
E. -2z^2 + 1
PS14031.02
A. z + 1
B. z^2 + 1
C. -z^2 + 1
D. -z^2 + z + 1
E. -2z^2 + 1
PS14031.02
\(z + \frac{1 - 2z^2}{z} = w/z\)
=> \(z(z + \frac{1 - 2z^2}{z}) = w\)
=> \(w = z^2 + (1 - 2z^2)\)
=> \(w = 1 - z^2\)
Answer C
General Discussion
28 Apr 2020, 08:38
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Bookmarks
gmatt1476
If z ≠ 0 and \(z + \frac{1 - 2z^2}{z} = \frac{w}{z}\), then \(w =\)
A. z + 1
B. z^2 + 1
C. -z^2 + 1
D. -z^2 + z + 1
E. -2z^2 + 1
PS14031.02
A. z + 1
B. z^2 + 1
C. -z^2 + 1
D. -z^2 + z + 1
E. -2z^2 + 1
PS14031.02
Given: \(z + \frac{1 - 2z^2}{z} = \frac{w}{z}\)
Eliminate the fractions, multiply both sides of the equation by \(z\) to get: \(z^2 + (1 - 2z^2) = w\)
Simplify to get: \(1 - z^2 = w\)
Rewrite as: \(-z^2 + 1= w\)
Answer: C
Cheers,
Brent
