Question
I'm studying for my math final and need help with a problem about three numbers X, Y and Z in the ratio 2:7:8. If 12 is subtracted from Y, the numbers form a geometric sequence. Can you find the values of X, Y and Z and confirm if there are multiple solutions to this problem?
Answers
GPT-4o mini
To solve for the values of \(X\), \(Y\), and \(Z\) based on the given ratio 2:7:8, we can establish the following relationships:
Let \(X = 2k\), \(Y = 7k\), and \(Z = 8k\) for some positive constant \(k\).
We know from the problem that if 12 is subtracted from \(Y\), the numbers will form a geometric sequence. The numbers then become \(X\), \(Y - 12\), and \(Z\), or \(2k\), \(7k - 12\), and \(8k\).
For the numbers to be in a geometric progression, the square of the middle term must equal the product of the first and last terms:
\[
(7k - 12)^2 = (2k)(8k)
\]
Expanding both sides:
\[
(7k - 12)^2 = 49k^2 - 168k + 144
\]
\[
(2k)(8k) = 16k^2
\]
Setting the equations equal gives:
\[
49k^2 - 168k + 144 = 16k^2
\]
Bringing all terms to one side:
\[
49k^2 - 16k^2 - 168k + 144 = 0
\]
\[
33k^2 - 168k + 144 = 0
\]
Next, we can use the quadratic formula \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Where \(a = 33\), \(b = -168\), and \(c = 144\):
Calculating the discriminant:
\[
b^2 - 4ac = (-168)^2 - 4 \cdot 33 \cdot 144 = 28224 - 19008 = 9216
\]
Now we compute \(k\):
\[
k = \frac{168 \pm \sqrt{9216}}{2 \cdot 33}
\]
Since \(\sqrt{9216} = 96\):
\[
k = \frac{168 \pm 96}{66}
\]
Calculating the two cases:
1. \(k = \frac{264}{66} = 4\)
2. \(k = \frac{72}{66} = \frac{12}{11}\)
Now, we have two values for \(k\):
1. If \(k = 4\):
\[
X = 2k = 8, \quad Y = 7k = 28, \quad Z = 8k = 32
\]
2. If \(k = \frac{12}{11}\):
\[
X = 2k = \frac{24}{11}, \quad Y = 7k = \frac{84}{11}, \quad Z = 8k = \frac{96}{11}
\]
Thus, we find:
1. For \(k = 4\): \(X = 8\), \(Y = 28\), \(Z = 32\)
2. For \(k = \frac{12}{11}\): \(X = \frac{24}{11}\), \(Y = \frac{84}{11}\), \(Z = \frac{96}{11}\)
Both solutions satisfy the conditions set in the problem.
**Final Values:**
1. \(X = 8\), \(Y = 28\), \(Z = 32\)
2. \(X = \frac{24}{11}\), \(Y = \frac{84}{11}\), \(Z = \frac{96}{11}\)
Thus, **there are multiple solutions to this problem**.
Let \(X = 2k\), \(Y = 7k\), and \(Z = 8k\) for some positive constant \(k\).
We know from the problem that if 12 is subtracted from \(Y\), the numbers will form a geometric sequence. The numbers then become \(X\), \(Y - 12\), and \(Z\), or \(2k\), \(7k - 12\), and \(8k\).
For the numbers to be in a geometric progression, the square of the middle term must equal the product of the first and last terms:
\[
(7k - 12)^2 = (2k)(8k)
\]
Expanding both sides:
\[
(7k - 12)^2 = 49k^2 - 168k + 144
\]
\[
(2k)(8k) = 16k^2
\]
Setting the equations equal gives:
\[
49k^2 - 168k + 144 = 16k^2
\]
Bringing all terms to one side:
\[
49k^2 - 16k^2 - 168k + 144 = 0
\]
\[
33k^2 - 168k + 144 = 0
\]
Next, we can use the quadratic formula \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Where \(a = 33\), \(b = -168\), and \(c = 144\):
Calculating the discriminant:
\[
b^2 - 4ac = (-168)^2 - 4 \cdot 33 \cdot 144 = 28224 - 19008 = 9216
\]
Now we compute \(k\):
\[
k = \frac{168 \pm \sqrt{9216}}{2 \cdot 33}
\]
Since \(\sqrt{9216} = 96\):
\[
k = \frac{168 \pm 96}{66}
\]
Calculating the two cases:
1. \(k = \frac{264}{66} = 4\)
2. \(k = \frac{72}{66} = \frac{12}{11}\)
Now, we have two values for \(k\):
1. If \(k = 4\):
\[
X = 2k = 8, \quad Y = 7k = 28, \quad Z = 8k = 32
\]
2. If \(k = \frac{12}{11}\):
\[
X = 2k = \frac{24}{11}, \quad Y = 7k = \frac{84}{11}, \quad Z = 8k = \frac{96}{11}
\]
Thus, we find:
1. For \(k = 4\): \(X = 8\), \(Y = 28\), \(Z = 32\)
2. For \(k = \frac{12}{11}\): \(X = \frac{24}{11}\), \(Y = \frac{84}{11}\), \(Z = \frac{96}{11}\)
Both solutions satisfy the conditions set in the problem.
**Final Values:**
1. \(X = 8\), \(Y = 28\), \(Z = 32\)
2. \(X = \frac{24}{11}\), \(Y = \frac{84}{11}\), \(Z = \frac{96}{11}\)
Thus, **there are multiple solutions to this problem**.