Q. \(x^{2}-2x-48=0\)

Q: How do you factor x^2-2x-48=0?

Find two numbers with product -48 and sum -2: -8 and 6. So x^2-2x-48=(x-8)(x+6).

Q: What are the solutions using factoring?

Set each factor to zero: x-8=0 gives x=8, and x+6=0 gives x=-6.

Q: How do you solve it using the quadratic formula?

With a=1, b=-2, c=-48: \displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-48)}}{2(1)}=\frac{2\pm\sqrt{196}}{2}=\frac{2\pm14}{2}.

Q: What is the discriminant b^2-4ac?

\Delta=(-2)^2-4(1)(-48)=4+192=196. Since \Delta>0, there are two real solutions.

Q: Can you check the solutions in the original equation?

For x=8: 64-16-48=0. For x=-6: 36+12-48=0. Both satisfy the equation.

Q: What is the sum and product of the roots?

For x^2-2x-48=0, sum = -b/a=2, product = c/a=-48. Indeed, 8+(-6)=2 and 8\cdot(-6)=-48.

Q: How do you solve by completing the square?

Rewrite as x^2-2x=48. Complete the square: (x-1)^2-1=48, so (x-1)^2=49, giving x-1=\pm7 and x=8,-6.

Answer

We solve the quadratic equation \(x^2-2x-48=0\) by factoring.

\[
x^2-2x-48=(x-8)(x+6)=0
\]

So \(x-8=0\) or \(x+6=0\).

\[
x=8 \quad \text{or} \quad x=-6
\]

Final answers: \(x=8\) or \(x=-6\).

Detailed Explanation

We want to solve the equation

\[
x^2 – 2x – 48 = 0
\]

Step 1: Factor the quadratic

To factor \(x^2 – 2x – 48\), we look for numbers \(a\) and \(b\) such that:

\[
a \cdot b = -48
\]

and

\[
a + b = -2
\]

Step 2: Choose numbers that multiply to \(-48\) and add to \(-2\)

Check factor pairs of \(-48\):

\[
-6 \cdot 8 = -48
\]

Now check the sum:

\[
-6 + 8 = 2 \quad \text{(not } -2\text{)}
\]

Try:

\[
6 \cdot (-8) = -48
\]

Now check the sum:

\[
6 + (-8) = -2 \quad \text{(perfect)}
\]

Step 3: Write the factorization

Using \(6\) and \(-8\), we factor:

\[
x^2 – 2x – 48 = (x + 6)(x – 8)
\]

Step 4: Set each factor equal to zero

If \((x + 6)(x – 8) = 0\), then either:

\[
x + 6 = 0
\]

\[
x – 8 = 0
\]

Step 5: Solve each simple equation

First equation:

\[
x + 6 = 0
\]

Subtract \(6\) from both sides:

\[
x = -6
\]

Second equation:

\[
x – 8 = 0
\]

Add \(8\) to both sides:

\[
x = 8
\]

Final Answer

The solutions to \(x^2 – 2x – 48 = 0\) are:

\[
x = -6 \quad \text{or} \quad x = 8
\]

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Algebra FAQ

How do you factor \(x^2-2x-48=0\)?

Find two numbers with product \(-48\) and sum \(-2\): \(-8\) and \(6\). So \(x^2-2x-48=(x-8)(x+6)\).

What are the solutions using factoring?

Set each factor to zero: \(x-8=0\) gives \(x=8\), and \(x+6=0\) gives \(x=-6\).

How do you solve it using the quadratic formula?

With \(a=1\), \(b=-2\), \(c=-48\): \(\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-48)}}{2(1)}=\frac{2\pm\sqrt{196}}{2}=\frac{2\pm14}{2}\).

What is the discriminant \(b^2-4ac\)?

\(\Delta=(-2)^2-4(1)(-48)=4+192=196\). Since \(\Delta>0\), there are two real solutions.

Can you check the solutions in the original equation?

For \(x=8\): \(64-16-48=0\). For \(x=-6\): \(36+12-48=0\). Both satisfy the equation.

What is the sum and product of the roots?

For \(x^2-2x-48=0\), sum \(= -b/a=2\), product \(= c/a=-48\). Indeed, \(8+(-6)=2\) and \(8\cdot(-6)=-48\).

How do you solve by completing the square?

Rewrite as \(x^2-2x=48\). Complete the square: \((x-1)^2-1=48\), so \((x-1)^2=49\), giving \(x-1=\pm7\) and \(x=8,-6\).

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