Q. \[ x^2 – 3x + 2 = 0 \]

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

Factor x^2-3x+2=(x-1)(x-2). Set each factor to zero: x-1=0 or x-2=0. So x=1 or x=2.

Q: How do you use the quadratic formula for x^2-3x+2=0?

Here a=1, b=-3, c=2. x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{3\pm\sqrt{9-8}}{2}=\frac{3\pm1}{2}. So x=1 or x=2.

Q: What is the discriminant, and what does it imply?

Discriminant \Delta=b^2-4ac=(-3)^2-4(1)(2)=9-8=1. Since \Delta>0, there are two distinct real solutions.

Q: Are there quick roots-checks for possible integer solutions?

Try factors of c=2: \pm1,\pm2. Test x=1: 1-3+2=0. Test x=2: 4-6+2=0. Hence the roots are 1 and 2.

Q: How do you find the sum and product of the roots?

For x^2-3x+2=0, sum =3 and product =2. Numbers with these properties satisfy r_1+r_2=3 and r_1r_2=2, giving r_1=1, r_2=2.

Q: Can you solve it by completing the square?

x^2-3x+2=0\Rightarrow x^2-3x=-2. Complete square: x^2-3x+\frac{9}{4}=\frac{1}{4}. So (x-\frac{3}{2})^2=\frac{1}{4}. Then x-\frac{3}{2}=\pm\frac{1}{2}, so x=1 or x=2.

Answer

We solve the quadratic \(x^2-3x+2=0\) by factoring.

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

So \(x-1=0\) or \(x-2=0\).

\[
x=1 \quad \text{or} \quad x=2
\]

Detailed Explanation

We want to solve the equation

\[
x^2 – 3x + 2 = 0
\]

Step 1: Factor the quadratic.

To factor \(x^2 – 3x + 2\), we need two numbers that:

Multiply to \(2\)
Add to \(-3\)

The factors of \(2\) are \(1\) and \(2\) (with signs). The pair that adds to \(-3\) is \(-1\) and \(-2\), because:

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

So the quadratic factors as:

\[
x^2 – 3x + 2 = (x – 1)(x – 2)
\]

Step 2: Set each factor equal to zero.

If a product is zero, then at least one factor must be zero:

\[
(x – 1)(x – 2) = 0
\]
\[
x – 1 = 0 \quad \text{or} \quad x – 2 = 0
\]

Step 3: Solve each simple equation.

First:

\[
x – 1 = 0
\]
\[
x = 1
\]

\[
x – 2 = 0
\]
\[
x = 2
\]

Final Answer.

The solutions to \(x^2 – 3x + 2 = 0\) are:

\[
x = 1 \quad \text{and} \quad x = 2
\]

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Algebra FAQ

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

Factor \(x^2-3x+2=(x-1)(x-2)\). Set each factor to zero: \(x-1=0\) or \(x-2=0\). So \(x=1\) or \(x=2\).

How do you use the quadratic formula for \(x^2-3x+2=0\)?

Here \(a=1\), \(b=-3\), \(c=2\). \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{3\pm\sqrt{9-8}}{2}=\frac{3\pm1}{2}\). So \(x=1\) or \(x=2\).

What is the discriminant, and what does it imply?

Discriminant \(\Delta=b^2-4ac=(-3)^2-4(1)(2)=9-8=1\). Since \(\Delta>0\), there are two distinct real solutions.

Are there quick roots-checks for possible integer solutions?

Try factors of \(c=2\): \(\pm1,\pm2\). Test \(x=1\): \(1-3+2=0\). Test \(x=2\): \(4-6+2=0\). Hence the roots are \(1\) and \(2\).

How do you find the sum and product of the roots?

For \(x^2-3x+2=0\), sum \(=3\) and product \(=2\). Numbers with these properties satisfy \(r_1+r_2=3\) and \(r_1r_2=2\), giving \(r_1=1\), \(r_2=2\).

Can you solve it by completing the square?

\(x^2-3x+2=0\Rightarrow x^2-3x=-2\). Complete square: \(x^2-3x+\frac{9}{4}=\frac{1}{4}\). So \((x-\frac{3}{2})^2=\frac{1}{4}\). Then \(x-\frac{3}{2}=\pm\frac{1}{2}\), so \(x=1\) or \(x=2\).

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