Q. \(x^2 – 5x – 24 = 0\)

Problem: Solve the quadratic equation

\[
x^2 – 5x – 24 = 0
\]

Step 1: Factor the quadratic.

To factor a quadratic of the form

\[
x^2 + bx + c
\]

we need two numbers that multiply to \(c\) and add to \(b\).

Here, \(b = -5\) and \(c = -24\).

So we look for numbers \(m\) and \(n\) such that:

\[
m \cdot n = -24
\]
\[
m + n = -5
\]

Step 2: Find the numbers.

List factor pairs of \(-24\):

\[
(-1, 24),\ (1, -24),\ (-2, 12),\ (2, -12),\ (-3, 8),\ (3, -8),\ (-4, 6),\ (4, -6)
\]

Check which pair adds to \(-5\):

\[
(-3) + (-8) = -11 \quad \text{(not } -5\text{)}
\]

\[
(-4) + (-6) = -10 \quad \text{(not } -5\text{)}
\]

Try a pair that includes \(-8\) and \(3\):

\[
3 + (-8) = -5
\]

Now check the product:

\[
3 \cdot (-8) = -24
\]

So the correct numbers are \(3\) and \(-8\).

Step 3: Write the factors.

Since the numbers are \(3\) and \(-8\), we can factor the quadratic as

\[
x^2 – 5x – 24 = (x + 3)(x – 8)
\]

Step 4: Set each factor equal to zero.

Using the Zero Product Property: if

\[
(a)(b) = 0
\]

then

\[
a = 0 \quad \text{or} \quad b = 0
\]

So we set:

\[
x + 3 = 0
\]
\[
x – 8 = 0
\]

Step 5: Solve each equation.

First equation:

\[
x + 3 = 0
\]
\[
x = -3
\]

Second equation:

\[
x – 8 = 0
\]
\[
x = 8
\]

Final Answer:

\[
x = -3 \quad \text{or} \quad x = 8
\]