When asked his age, the algebra teacher said, " if you square my age, then subtract 23 times my age, the result is 50." How old is he?

Options

A. 23
B. 25
C. 27
D. 46

Answer

To solve the problem of the algebra teacher's age, we can follow these steps:

### Step 1: Define the variable
Let the age of the teacher be represented by \( x \).

### Step 2: Set up the equation
According to the problem, if you square his age and then subtract 23 times his age, the result is 50. This can be expressed mathematically as:
\[ x^2 - 23x = 50 \]

### Step 3: Rearrange the equation
To form a standard quadratic equation, we need to move all terms to one side:
\[ x^2 - 23x - 50 = 0 \]

### Step 4: Identify coefficients
In the standard form \( ax^2 + bx + c = 0 \), we can identify:
- \( a = 1 \)
- \( b = -23 \)
- \( c = -50 \)

### Step 5: Use the quadratic formula
The quadratic formula to find the roots of the equation is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ x = \frac{-(-23) \pm \sqrt{(-23)^2 - 4 \cdot 1 \cdot (-50)}}{2 \cdot 1} \]

### Step 6: Calculate the discriminant
First, calculate \( b^2 - 4ac \):
\[ (-23)^2 = 529 \]
\[ 4 \cdot 1 \cdot (-50) = -200 \quad \text{(which becomes positive when subtracted)} \]
Thus,
\[ b^2 - 4ac = 529 + 200 = 729 \]

### Step 7: Substitute back into the formula
Now substitute back into the quadratic formula:
\[ x = \frac{23 \pm \sqrt{729}}{2} \]

### Step 8: Calculate the square root
The square root of 729 is:
\[ \sqrt{729} = 27 \]
So,
\[ x = \frac{23 \pm 27}{2} \]

### Step 9: Solve for the two possible values of \( x \)
Calculating the two possible solutions:
1. \( x = \frac{23 + 27}{2} = \frac{50}{2} = 25 \)
2. \( x = \frac{23 - 27}{2} = \frac{-4}{2} = -2 \)

### Step 10: Determine the valid age
Since age cannot be negative, we discard \( x = -2 \). Therefore, the teacher's age is:
\[ x = 25 \]

### Conclusion
The age of the algebra teacher is **25 years**.

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