Q. \(x^{2}-3x+4\)

Q: What is the value of x^{2}-3x+4 at x=0?

Substitute x=0: 0^{2}-3(0)+4=4.

Q: Factor x^{2}-3x+4 if possible.

Discriminant b^{2}-4ac=9-16=-7<0, so it does not factor over real numbers.

Q: Complete the square for x^{2}-3x+4.

x^{2}-3x+4=\left(x-\frac{3}{2}\right)^{2}+\frac{7}{4}.

Q: Find the minimum value of x^{2}-3x+4 and the x-coordinate where it occurs.

Vertex at x=\frac{3}{2}. Minimum value is \frac{7}{4}.

Q: Solve x^{2}-3x+4=0 over the complex numbers.

x=\frac{3\pm\sqrt{-7}}{2}=\frac{3\pm i\sqrt{7}}{2}.

Q: Determine whether the expression is always positive.

Since x^{2}-3x+4=\left(x-\frac{3}{2}\right)^{2}+\frac{7}{4} and \frac{7}{4}>0, it is always positive.

Answer

Factor the quadratic \(x^2-3x+4\) if possible. Its discriminant is

\[
\Delta=b^2-4ac=(-3)^2-4(1)(4)=9-16=-7<0 \]

Since \(\Delta\) is negative, it has no real (or rational) linear factors, so it cannot be factored over the real numbers. The quadratic is already the simplified form.

\(\boxed{x^2-3x+4}\)

Detailed Explanation

We are asked to work with the expression

\[
x^{2}-3x+4
\]

This is a polynomial in standard form. The most common “step-by-step” task in this situation is to check whether it can be factored.

\nStep 1: Identify coefficients

The expression has the form \(ax^{2}+bx+c\), where:

\[
a=1,\quad b=-3,\quad c=4
\]

\nStep 2: Try factoring into two binomials

Suppose we can write:

\[
x^{2}-3x+4=(x+m)(x+n)
\]

Expanding the right-hand side gives:

\[
(x+m)(x+n)=x^{2}+(m+n)x+mn
\]

Match this with \(x^{2}-3x+4\). Then we need:

\[
m+n=-3
\]

\[
mn=4
\]

\nStep 3: Find numbers \(m\) and \(n\)

We need two numbers whose product is \(4\) and whose sum is \(-3\).

List factor pairs of \(4\):

\[
1\cdot 4=4,\quad 2\cdot 2=4,\quad (-1)\cdot(-4)=4,\quad (-2)\cdot(-2)=4
\]

Now check their sums:

\[
1+4=5
\]

\[
2+2=4
\]

\[
(-1)+(-4)=-5
\]

\[
(-2)+(-2)=-4
\]

None of these sums equal \(-3\). Therefore, the polynomial does not factor nicely over the integers.

\nStep 4: Confirm the polynomial is already simplified

Since it cannot be factored into integer binomials, the given expression is already in simplest form (as a quadratic polynomial).

\nFinal Answer

\[
x^{2}-3x+4
\]

See full solution

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Algebra FAQ

What is the value of \(x^{2}-3x+4\) at \(x=0\)?

Substitute \(x=0\): \(0^{2}-3(0)+4=4\).

Factor \(x^{2}-3x+4\) if possible.

Discriminant \(b^{2}-4ac=9-16=-7<0\), so it does not factor over real numbers.

Complete the square for \(x^{2}-3x+4\).

\(x^{2}-3x+4=\left(x-\frac{3}{2}\right)^{2}+\frac{7}{4}\).

Find the minimum value of \(x^{2}-3x+4\) and the \(x\)-coordinate where it occurs.

Vertex at \(x=\frac{3}{2}\). Minimum value is \(\frac{7}{4}\).

Solve \(x^{2}-3x+4=0\) over the complex numbers.

\(x=\frac{3\pm\sqrt{-7}}{2}=\frac{3\pm i\sqrt{7}}{2}\).

Determine whether the expression is always positive.

Since \(x^{2}-3x+4=\left(x-\frac{3}{2}\right)^{2}+\frac{7}{4}\) and \(\frac{7}{4}>0\), it is always positive.

Solve x^(2)-3x+4 step by step.
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