Q. \(x^2 – 2x – 15 = 0\)

Q: What factors solve x^2-2x-15=0?

Factor: x^2-2x-15=(x-5)(x+3). Set (x-5)(x+3)=0, so x=5 or x=-3.

Q: Use the quadratic formula for x^2-2x-15=0.

For a=1,b=-2,c=-15: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2\pm\sqrt{4+60}}{2}=\frac{2\pm 8}{2}. Thus x=5 or x=-3.

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

b^2-4ac=(-2)^2-4(1)(-15)=4+60=64. Since it is positive, there are two real roots.

Q: How do you complete the square?

x^2-2x-15=0 \Rightarrow x^2-2x=15. Add 1: (x-1)^2=16. So x-1=\pm 4, giving x=5 or x=-3.

Q: How can you check the solutions quickly?

Substitute x=5: 25-10-15=0. Substitute x=-3: 9+6-15=0. Both satisfy the equation.

Q: What are the roots’ sum and product?

For x^2-2x-15=0: sum = -\frac{b}{a}=2. product =\frac{c}{a}=-15. Roots 5 and -3 match: 5+(-3)=2, 5\cdot(-3)=-15.

Answer

We solve the quadratic \(x^2-2x-15=0\) by factoring:

\[
x^2-2x-15=(x-5)(x+3)=0
\]

So \(x-5=0\) or \(x+3=0\).

\[
x=5 \quad \text{or} \quad x=-3
\]

Final answer: \(x=5\) or \(x=-3\).

Detailed Explanation

We need to solve the quadratic equation

\[
x^2 – 2x – 15 = 0
\]

Step 1: Factor the quadratic

We want to write the left-hand side in factored form:

\[
x^2 – 2x – 15 = (x + a)(x + b)
\]

When we expand \((x + a)(x + b)\), we get

\[
(x + a)(x + b) = x^2 + (a+b)x + ab
\]

Match coefficients with \(x^2 – 2x – 15\):

\[
a + b = -2
\]
\[
ab = -15
\]

Step 2: Find numbers that work

We need two numbers whose product is \(-15\) and whose sum is \(-2\). The pair \(-5\) and \(3\) works because

\[
(-5)(3) = -15
\]
\[
(-5) + 3 = -2
\]

So we factor the quadratic as:

\[
x^2 – 2x – 15 = (x – 5)(x + 3)
\]

Step 3: Set each factor equal to zero

\[
(x – 5)(x + 3) = 0
\]

then either

\[
x – 5 = 0
\]
or
\[
x + 3 = 0
\]

Step 4: Solve each simple equation

First equation:

\[
x – 5 = 0
\]
\[
x = 5
\]

Second equation:

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

Final Answer

The solutions to

\[
x^2 – 2x – 15 = 0
\]

are

\[
x = 5 \quad \text{or} \quad x = -3
\]

See full solution

Graph

Need help with homework? Try our AI solutions now!

Homework AI

Algebra FAQ

What factors solve \(x^2-2x-15=0\)?

Factor: \(x^2-2x-15=(x-5)(x+3)\). Set \((x-5)(x+3)=0\), so \(x=5\) or \(x=-3\).

Use the quadratic formula for \(x^2-2x-15=0\).

For \(a=1,b=-2,c=-15\): \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2\pm\sqrt{4+60}}{2}=\frac{2\pm 8}{2}\). Thus \(x=5\) or \(x=-3\).

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

\(b^2-4ac=(-2)^2-4(1)(-15)=4+60=64\). Since it is positive, there are two real roots.

How do you complete the square?

\(x^2-2x-15=0 \Rightarrow x^2-2x=15\). Add 1: \((x-1)^2=16\). So \(x-1=\pm 4\), giving \(x=5\) or \(x=-3\).

How can you check the solutions quickly?

Substitute \(x=5\): \(25-10-15=0\). Substitute \(x=-3\): \(9+6-15=0\). Both satisfy the equation.

What are the roots’ sum and product?

For \(x^2-2x-15=0\): sum \(= -\frac{b}{a}=2\). product \(=\frac{c}{a}=-15\). Roots \(5\) and \(-3\) match: \(5+(-3)=2\), \(5\cdot(-3)=-15\).

Use this to solve the equation.
Add practice with math AI tools.

298,376+ active customers
Math, Geometry, Trigonometry, etc.

Math Solver Geometry Solver Trigonometry Solver