Let the original price of each computer be \( P \). The retailer planned to buy \( N \) computers.
The total amount the retailer planned to spend is \( 1,800,000 \), so we have the equation:
\[ N \times P = 1,800,000 \]
After the price reduction of \( 4,000 \), the new price of each computer becomes \( P - 4,000 \). With this new price, the retailer is able to buy \( N + 5 \) computers for the same total budget. So we can set up a second equation:
\[ (N + 5) \times (P - 4,000) = 1,800,000 \]
Now we can simplify the first equation to express \( P \):
\[ P = \frac{1,800,000}{N} \]
Substituting \( P \) in the second equation, we get:
\[ (N + 5) \left( \frac{1,800,000}{N} - 4,000 \right) = 1,800,000 \]
Expanding this, we have:
\[ (N + 5) \left( \frac{1,800,000 - 4,000N}{N} \right) = 1,800,000 \]
Multiplying both sides by \( N \) to eliminate the fraction:
\[ (N + 5)(1,800,000 - 4,000N) = 1,800,000N \]
Expanding the left side:
\[ 1,800,000N + 9,000,000 - 4,000N^2 - 20,000N = 1,800,000N \]
Now, we can subtract \( 1,800,000N \) from both sides:
\[ 9,000,000 - 4,000N^2 - 20,000N = 0 \]
Rearranging gives us a standard quadratic form:
\[ 4,000N^2 + 20,000N - 9,000,000 = 0 \]
To simplify, we can divide the entire equation by 4,000:
\[ N^2 + 5N - 2,250 = 0 \]
Next, we can use the quadratic formula \( N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = 5 \), and \( c = -2,250 \):
First calculate the discriminant:
\[ b^2 - 4ac = 5^2 - 4 \times 1 \times (-2,250) = 25 + 9,000 = 9,025 \]
Now, taking the square root:
\[ \sqrt{9,025} = 95 \]
Now, substituting back into the quadratic formula:
\[ N = \frac{-5 \pm 95}{2} \]
Calculating the two possible values:
- \( N = \frac{90}{2} = 45 \)
- \( N = \frac{-100}{2} = -50 \) (not a valid solution)
Thus, the retailer originally planned to buy \( N = 45 \) computers.
Finally, the number of computers the retailer actually bought after the price reduction is:
\[ N + 5 = 45 + 5 = 50 \]
So, the retailer bought 50 computers.
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