To solve the problem of finding the numbers whose product with 3 less than themselves equals 378, we can follow these systematic steps:
1. Define the problem with a variable:
Let the number be \( x \). According to the problem, the product of the number and 3 less than the number is 378. So, we write:
[tex]\[
x \times (x - 3) = 378
\][/tex]
2. Formulate the equation:
The above expression simplifies to a standard quadratic equation:
[tex]\[
x^2 - 3x - 378 = 0
\][/tex]
3. Use the quadratic formula:
The quadratic formula for solving \( ax^2 + bx + c = 0 \) is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, \( a = 1 \), \( b = -3 \), and \( c = -378 \).
4. Calculate the discriminant:
The discriminant (\( \Delta \)) is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substituting the values:
[tex]\[
\Delta = (-3)^2 - 4 \times 1 \times (-378) = 9 + 1512 = 1521
\][/tex]
5. Find the square root of the discriminant:
[tex]\[
\sqrt{1521} = 39
\][/tex]
6. Calculate the two possible solutions:
Substituting the discriminant and other values back into the quadratic formula:
[tex]\[
x_1 = \frac{-(-3) + 39}{2 \times 1} = \frac{3 + 39}{2} = \frac{42}{2} = 21
\][/tex]
[tex]\[
x_2 = \frac{-(-3) - 39}{2 \times 1} = \frac{3 - 39}{2} = \frac{-36}{2} = -18
\][/tex]
7. Check for positivity:
Since we are asked to assume both numbers are positive, we discard \( x_2 \) (which is \(-18\)) because it is negative.
8. Conclusion:
The solution to the problem is that the positive number is \( x = 21 \). Therefore, the two numbers in question are:
[tex]\[
21 \quad \text{and} \quad (21-3) = 18
\][/tex]
Hence, the numbers are 21 and 18.
