Q. \(x^2-7x+12\)

Q: What are the roots (solutions) of x^2-7x+12=0?

Solving (x-3)(x-4)=0 gives x=3 or x=4.

Q: What is the discriminant of x^2-7x+12?

For ax^2+bx+c, D=b^2-4ac=(-7)^2-4(1)(12)=49-48=1.

Q: Can you complete the square for x^2-7x+12?

x^2-7x+12=(x-\tfrac{7}{2})^2-\tfrac{1}{4}. Then set equal to 0 if needed.

Q: How do you check the factors by multiplying?

(x-3)(x-4)=x^2-4x-3x+12=x^2-7x+12.

Q: What is the vertex of the parabola y=x^2-7x+12?

For y=ax^2+bx+c, x-coordinate is -\frac{b}{2a}=\frac{7}{2}. Then y= \left(\frac{7}{2}\right)^2-7\left(\frac{7}{2}\right)+12=\frac{-1}{4}.

Answer

Factor the quadratic by finding two numbers that multiply to \(12\) and add to \(-7\). Those numbers are \(-3\) and \(-4\).

\[
x^2-7x+12=(x-3)(x-4)
\]

So the solutions are \(x=3\) and \(x=4\).

Detailed Explanation

We want to factor the expression \(x^2 – 7x + 12\). Factoring means rewriting it as a product of two simpler polynomials.

Step 1: Identify the form

The expression has the form

\[
x^2 – 7x + 12
\]
This matches the general quadratic form \(x^2 + bx + c\), where:

\[
b = -7, \quad c = 12
\]

Step 2: Find two numbers

We need two integers \(m\) and \(n\) such that:

\[
m \cdot n = 12
\]
and

\[
m + n = -7
\]

Step 3: List factor pairs of 12

Possible pairs \((m, n)\) whose product is \(12\) are:

\[
(1, 12),\ (2, 6),\ (3, 4)
\]
To get a sum of \(-7\), both numbers must be negative, so we check:

\[
(-3, -4)
\]

Step 4: Verify the condition

Check the product:

\[
(-3)(-4) = 12
\]
Check the sum:

\[
-3 + (-4) = -7
\]
Both conditions are satisfied.

Step 5: Rewrite using the found numbers

Now we can factor the quadratic as:

\[
x^2 – 7x + 12 = (x – 3)(x – 4)
\]

Final Answer

\[
x^2 – 7x + 12 = (x – 3)(x – 4)
\]

See full solution

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Algebra FAQ

How do you factor \(x^2-7x+12\)?

\(\,x^2-7x+12=(x-3)(x-4)\).

What are the roots (solutions) of \(x^2-7x+12=0\)?

Solving \((x-3)(x-4)=0\) gives \(x=3\) or \(x=4\).

What is the discriminant of \(x^2-7x+12\)?

For \(ax^2+bx+c\), \(D=b^2-4ac=(-7)^2-4(1)(12)=49-48=1\).

Can you complete the square for \(x^2-7x+12\)?

\(x^2-7x+12=(x-\tfrac{7}{2})^2-\tfrac{1}{4}\). Then set equal to \(0\) if needed.

How do you check the factors by multiplying?

\((x-3)(x-4)=x^2-4x-3x+12=x^2-7x+12\).

What is the vertex of the parabola \(y=x^2-7x+12\)?

For \(y=ax^2+bx+c\), \(x\)-coordinate is \(-\frac{b}{2a}=\frac{7}{2}\). Then \(y= \left(\frac{7}{2}\right)^2-7\left(\frac{7}{2}\right)+12=\frac{-1}{4}\).

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