Q. \(x^2 – 5x – 14\)

Q: Solve x^2-5x-14=0.

From (x-7)(x+2)=0, x=7 or x=-2.

Q: Find the x-intercepts.

x-intercepts are (7,0) and (-2,0).

Q: Determine the vertex of y=x^2-5x-14.

Vertex x-value is x=-\frac{b}{2a}=\frac{5}{2}. Then y=\left(\frac{5}{2}\right)^2-5\left(\frac{5}{2}\right)-14=-\frac{57}{4}.

Q: Is y=x^2-5x-14 positive or negative between roots?

Since the parabola opens up, it is negative between x=-2 and x=7. Positive outside that interval.

Answer

Factor the quadratic.

\[
x^2 – 5x – 14 = (x – 7)(x + 2)
\]

So the solutions are when each factor is zero:

\[
x – 7 = 0 \Rightarrow x = 7,\qquad x + 2 = 0 \Rightarrow x = -2
\]

Final result: \(x = 7\) or \(x = -2\).

Detailed Explanation

We are asked to work with the expression

\[
x^2 – 5x – 14
\]

A common goal with a quadratic expression like this is to factor it into two binomials, if possible.

Step 1: Identify the numbers to factor.

For a quadratic of the form

\[
x^2 + bx + c
\]

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

Here:

\(b = -5\)
\(c = -14\)

Step 2: Find two numbers whose product is \(-14\).

We need numbers \(m\) and \(n\) such that

\[
mn = -14
\]

List factor pairs of \(-14\):

\(1 \cdot -14 = -14\)
\(-1 \cdot 14 = -14\)
\(2 \cdot -7 = -14\)
\(-2 \cdot 7 = -14\)

Step 3: Check which pair adds to \(-5\).

We also need

\[
m + n = -5
\]

Check each pair:

\(1 + (-14) = -13\)
\((-1) + 14 = 13\)
\(2 + (-7) = -5\) ✓
\((-2) + 7 = 5\)

The correct pair is \(2\) and \(-7\).

Step 4: Write the factored form.

Since the product is \(-14\) and the sum is \(-5\), the quadratic factors as

\[
x^2 – 5x – 14 = (x + 2)(x – 7)
\]

Final Answer:

\[
x^2 – 5x – 14 = (x + 2)(x – 7)
\]

See full solution

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Algebra FAQ

Factorize \(x^2-5x-14\).

\(x^2-5x-14=(x-7)(x+2)\).

Solve \(x^2-5x-14=0\).

From \((x-7)(x+2)=0\), \(x=7\) or \(x=-2\).

What are the roots of \(x^2-5x-14\)?

The roots are \(7\) and \(-2\).

Find the \(x\)-intercepts.

\(x\)-intercepts are \((7,0)\) and \((-2,0)\).

Determine the vertex of \(y=x^2-5x-14\).

Vertex \(x\)-value is \(x=-\frac{b}{2a}=\frac{5}{2}\). Then \(y=\left(\frac{5}{2}\right)^2-5\left(\frac{5}{2}\right)-14=-\frac{57}{4}\).

Is \(y=x^2-5x-14\) positive or negative between roots?

Since the parabola opens up, it is negative between \(x=-2\) and \(x=7\). Positive outside that interval.

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