Q. \(x^2-2x-3=0\)

Q: . Solve x^2-2x-3=0 by factoring.

. Solve x^2-2x-3=(x-3)(x+1)=0. Thus x=3 or x=-1.

Q: . Use the quadratic formula to solve x^2-2x-3=0.

. With a=1,b=-2,c=-3: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2\pm\sqrt{4+12}}{2}=1\pm 2. So x=3 or x=-1.

Q: . What is the discriminant of x^2-2x-3=0, and what does it mean?

. D=b^2-4ac=(-2)^2-4(1)(-3)=4+12=16. Since D>0, there are two distinct real solutions: x=3,-1.

Q: . Complete the square for x^2-2x-3=0.

. x^2-2x-3=0\Rightarrow x^2-2x=3\Rightarrow (x-1)^2-1=3\Rightarrow (x-1)^2=4. So x=1\pm 2\Rightarrow x=3,-1.

Q: . What are the roots’ sum and product?

. For x^2-2x-3=0, sum = -\frac{b}{a}=2, product =\frac{c}{a}=-3. Roots 3 and -1 match: 3+(-1)=2, 3\cdot(-1)=-3.

Q: . Check the solutions x=3 and x=-1 in the original equation.

. 3^2-2(3)-3=9-6-3=0. Also (-1)^2-2(-1)-3=1+2-3=0. Both satisfy the equation.

Answer

We solve the quadratic equation by factoring:

\[
x^2-2x-3=0
\]

Find two numbers with product \(-3\) and sum \(-2\), which are \(-3\) and \(1\). So:

\[
x^2-2x-3=(x-3)(x+1)=0
\]

Set each factor equal to zero:

\[
x-3=0 \Rightarrow x=3
\]
\[
x+1=0 \Rightarrow x=-1
\]

Final result: \(x=3\) or \(x=-1\).

Detailed Explanation

We want to solve the equation

\[
x^2 – 2x – 3 = 0.
\]

Step 1: Identify the type of equation.

This is a quadratic equation because the highest power of \(x\) is \(2\). For a quadratic of the form

\[
ax^2 + bx + c = 0,
\]

we have:

\[
a = 1,\quad b = -2,\quad c = -3.
\]

Step 2: Use factoring (if possible).

We try to factor the quadratic into the form

\[
x^2 – 2x – 3 = (x + m)(x + n),
\]

where \(m\) and \(n\) are numbers.

Step 3: Match coefficients.

When we expand

\[
(x + m)(x + n),
\]

we get

\[
x^2 + (m+n)x + mn.
\]

So we need:

\[
m + n = -2
\]

and

\[
mn = -3.
\]

Step 4: Find numbers that multiply to \(-3\) and add to \(-2\).

The factor pairs of \(-3\) are:

\(\;1\) and \(-3\), or \(-1\) and \(3\).

Check which pair adds to \(-2\):

\(1 + (-3) = -2\) works.

So we take \(m = 1\) and \(n = -3\).

Step 5: Write the factored form.

\[
x^2 – 2x – 3 = (x+1)(x-3).
\]

Step 6: Set each factor equal to zero.

Because

\[
(x+1)(x-3)=0,
\]

the zero-product property tells us:

\[
x+1 = 0 \quad \text{or} \quad x-3 = 0.
\]

Step 7: Solve each equation.

First equation:

\[
x+1=0
\]

Subtract \(1\) from both sides:

\[
x=-1.
\]

Second equation:

\[
x-3=0
\]

Add \(3\) to both sides:

\[
x=3.
\]

Final Answer:

\[
x = -1 \quad \text{or} \quad x = 3.
\]

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Algebra FAQ

. Solve \(x^2-2x-3=0\) by factoring.

. Solve \(x^2-2x-3=(x-3)(x+1)=0\). Thus \(x=3\) or \(x=-1\).

. Use the quadratic formula to solve \(x^2-2x-3=0\).

. With \(a=1,b=-2,c=-3\): \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2\pm\sqrt{4+12}}{2}=1\pm 2\). So \(x=3\) or \(x=-1\).

. What is the discriminant of \(x^2-2x-3=0\), and what does it mean?

. \(D=b^2-4ac=(-2)^2-4(1)(-3)=4+12=16\). Since \(D>0\), there are two distinct real solutions: \(x=3,-1\).

. Complete the square for \(x^2-2x-3=0\).

. \(x^2-2x-3=0\Rightarrow x^2-2x=3\Rightarrow (x-1)^2-1=3\Rightarrow (x-1)^2=4\). So \(x=1\pm 2\Rightarrow x=3,-1\).

. What are the roots’ sum and product?

. For \(x^2-2x-3=0\), sum \(= -\frac{b}{a}=2\), product \(=\frac{c}{a}=-3\). Roots \(3\) and \(-1\) match: \(3+(-1)=2\), \(3\cdot(-1)=-3\).

. Check the solutions \(x=3\) and \(x=-1\) in the original equation.

. \(3^2-2(3)-3=9-6-3=0\). Also \((-1)^2-2(-1)-3=1+2-3=0\). Both satisfy the equation.

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