Q. \(x^2 – x – 2\)

Q: Solve x^2-x-2=0.

Factor x^2-x-2=(x-2)(x+1). So x=2 or x=-1.

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

a=1,b=-1,c=-2. Then x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-2)}}{2(1)}=\frac{1\pm\sqrt{9}}{2}, so x=2,-1.

Q: Find the roots of x^2-x-2 without factoring.

Compute the discriminant D=b^2-4ac=1-4(1)(-2)=9. Then x=\frac{-b\pm\sqrt{D}}{2a}=\frac{1\pm3}{2}, giving 2 and -1.

Q: What is the vertex of y=x^2-x-2?

For y=ax^2+bx+c, vertex x-coordinate is -\frac{b}{2a}=-\frac{-1}{2}= \frac{1}{2}. Then y(\tfrac12)=(\tfrac12)^2-\tfrac12-2=\tfrac14-\tfrac12-2=-\tfrac{9}{4}.

Q: Determine the x-intercepts of x^2-x-2.

x-intercepts occur when x^2-x-2=0. Thus intercepts are at x=2 and x=-1, giving points (2,0) and (-1,0).

Q: Is x^2-x-2 positive or negative for a given x?

Since x^2-x-2=(x-2)(x+1), the sign depends on factors. Between -1 and 2, the product is negative; outside that interval, it’s positive. At x=-1 and x=2, it equals 0.

Answer

We solve the quadratic \(x^2 – x – 2 = 0\) by factoring.

Find two numbers that multiply to \(-2\) and add to \(-1\): \(-2\) and \(1\).

\[
x^2 – x – 2 = (x – 2)(x + 1)
\]

Set each factor equal to \(0\):

\[
x – 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]

Final result: \(x = 2\) or \(x = -1\).

Detailed Explanation

We want to simplify and factor the expression \(x^2 – x – 2\).

Step 1: Identify the structure.

The expression \(x^2 – x – 2\) is a quadratic polynomial in standard form \(ax^2 + bx + c\), where:

\(a = 1\), \(b = -1\), and \(c = -2\).

Step 2: Factor using two numbers.

To factor \(x^2 – x – 2\), we look for two integers \(m\) and \(n\) such that:

\(m \cdot n = a \cdot c = 1 \cdot (-2) = -2\)

and

\(m + n = b = -1\).

Step 3: Find the integers.

The pairs of integers whose product is \(-2\) are:

\(2 \cdot (-1) = -2\)

and

\((-2) \cdot 1 = -2\).

Check the sums:

\(2 + (-1) = 1\) (not \(-1\))

\((-2) + 1 = -1\) (this matches).

Step 4: Write the factored form.

So we use \(-2\) and \(1\) to split the middle term:

\[
x^2 – x – 2 = x^2 + (-2 + 1)x – 2.
\]

Step 5: Factor by grouping.

Rewrite and group terms:

\[
x^2 – x – 2 = \bigl(x^2 – 2x\bigr) + \bigl(x – 2\bigr).
\]

Factor each group:

\[
x^2 – 2x = x(x – 2)
\]

\[
x – 2 = 1(x – 2).
\]

Now combine:

\[
\bigl(x(x – 2)\bigr) + \bigl(1(x – 2)\bigr) = (x – 2)(x + 1).
\]

Final Answer:

\[
x^2 – x – 2 = (x – 2)(x + 1).
\]

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Algebra FAQ

Solve \(x^2-x-2=0\).

Factor \(x^2-x-2=(x-2)(x+1)\). So \(x=2\) or \(x=-1\).

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

\(a=1,b=-1,c=-2\). Then \(x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-2)}}{2(1)}=\frac{1\pm\sqrt{9}}{2}\), so \(x=2,-1\).

Find the roots of \(x^2-x-2\) without factoring.

Compute the discriminant \(D=b^2-4ac=1-4(1)(-2)=9\). Then \(x=\frac{-b\pm\sqrt{D}}{2a}=\frac{1\pm3}{2}\), giving \(2\) and \(-1\).

What is the vertex of \(y=x^2-x-2\)?

For \(y=ax^2+bx+c\), vertex \(x\)-coordinate is \(-\frac{b}{2a}=-\frac{-1}{2}= \frac{1}{2}\). Then \(y(\tfrac12)=(\tfrac12)^2-\tfrac12-2=\tfrac14-\tfrac12-2=-\tfrac{9}{4}\).

Determine the \(x\)-intercepts of \(x^2-x-2\).

\(x\)-intercepts occur when \(x^2-x-2=0\). Thus intercepts are at \(x=2\) and \(x=-1\), giving points \((2,0)\) and \((-1,0)\).

Is \(x^2-x-2\) positive or negative for a given \(x\)?

Since \(x^2-x-2=(x-2)(x+1)\), the sign depends on factors. Between \(-1\) and \(2\), the product is negative; outside that interval, it’s positive. At \(x=-1\) and \(x=2\), it equals \(0\).

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