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16 Sep 2014, 01:53
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D
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Difficulty:
65%
(hard)
Question Stats:
54% (02:18) correct
46%
(02:38)
wrong
based on 72
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Given that \(x^4 - 25x^2 = -144\), which of the following is NOT a sum of two possible values of \(x\)?
A. -7
B. -1
C. 0
D. 3
E. 7
A. -7
B. -1
C. 0
D. 3
E. 7
16 Sep 2014, 01:53
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Official Solution:
Given that \(x^4 - 25x^2 = -144\), which of the following is NOT a sum of two possible values of \(x\)?
A. -7
B. -1
C. 0
D. 3
E. 7
First, solve the given equation:
\(x^4 - 25x^2 = -144\)
\(x^4 - 25x^2 + 144 = 0\)
Substitute \(z = x^2\):
\(z^2 - 25z + 144 = 0\)
Use trial and error to find a pair of factors of 144 that sum to 25. Start with 12 & 12, which sum to 24; then try 16 and 9, which work.
\((z - 16)(z - 9) = 0\)
\(z = 16\) or \(z = 9\)
Now put \(x^2\) back in:
\(x^2 = 16\) or \(x^2 = 9\)
\(x = -4\) or \(4\), or \(x = -3\) or \(3\). There are four possible solutions.
Possible sums of two solutions include the following:
\(-7 = -4 + -3\)
\(-1 = -4 + 3 \)
\(0 = -4 + 4\) or \(-3 + 3\)
\(7 = 4 + 3\)
However, 3 itself is NOT a possible sum of two solutions of the equation.
Answer: D
Given that \(x^4 - 25x^2 = -144\), which of the following is NOT a sum of two possible values of \(x\)?
A. -7
B. -1
C. 0
D. 3
E. 7
First, solve the given equation:
\(x^4 - 25x^2 = -144\)
\(x^4 - 25x^2 + 144 = 0\)
Substitute \(z = x^2\):
\(z^2 - 25z + 144 = 0\)
Use trial and error to find a pair of factors of 144 that sum to 25. Start with 12 & 12, which sum to 24; then try 16 and 9, which work.
\((z - 16)(z - 9) = 0\)
\(z = 16\) or \(z = 9\)
Now put \(x^2\) back in:
\(x^2 = 16\) or \(x^2 = 9\)
\(x = -4\) or \(4\), or \(x = -3\) or \(3\). There are four possible solutions.
Possible sums of two solutions include the following:
\(-7 = -4 + -3\)
\(-1 = -4 + 3 \)
\(0 = -4 + 4\) or \(-3 + 3\)
\(7 = 4 + 3\)
However, 3 itself is NOT a possible sum of two solutions of the equation.
Answer: D
11 Oct 2014, 11:42
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Bunuel
Curious, why is it that we cannot take the square root at \(x^4 - 25x^2 + 144 = 0\) and get \(x^2 - 25x + 12 = 0\)? Obviously something is wrong doing that... should we always look to substitute another variable into quadratic expressions with powers of 3, 4, 5 etc?
I understand that the math does not work, which is why its wrong, but perhaps I'm looking for a more technical explanation for these types of problems
edit: actually never mind altogether. quite a silly questions now that i think about it. O_o
