Q. \(x^2-8x+12=0\)

Q: Solve x^2-8x+12=0 by factoring.

Find numbers with product 12 and sum 8: 3 and 4. So x^2-8x+12=(x-4)(x-3)=0. Thus x=4 or x=3.

Q: Solve x^2-8x+12=0 using the quadratic formula.

a=1, b=-8, c=12. Then x=\frac{8\pm\sqrt{64-48}}{2}=\frac{8\pm\sqrt{16}}{2}=\frac{8\pm4}{2}. So x=6 or x=2. (Check: actually \frac{8\pm4}{2} \Rightarrow 6,2.) But factoring gives 3,4? Recheck: b^2-4ac=64-48=16. Correct roots are x=6,2.

Q: Verify the roots in x^2-8x+12=0.

For x=6: 36-48+12=0. For x=2: 4-16+12=0. Both satisfy the equation.

Q: What is the discriminant b^2-4ac and what does it mean?

D=(-8)^2-4(1)(12)=64-48=16. Since D>0, there are two distinct real solutions.

Q: Compute the sum and product of the roots.

Roots r_1,r_2 satisfy r_1+r_2=-\frac{b}{a}=8 and r_1r_2=\frac{c}{a}=12. With roots 6 and 2: sum 8, product 12.

Q: Solve the equation by completing the square.

x^2-8x+12=(x^2-8x+16)-4=(x-4)^2-4=0. Then (x-4)^2=4, so x-4=\pm2. Thus x=6 or x=2.

Answer

We solve the quadratic \(x^2-8x+12=0\) by factoring:

\[
x^2-8x+12=(x-2)(x-6)=0
\]

So \(x-2=0\) or \(x-6=0\).

Final result: \(x=2\) or \(x=6\).

Detailed Explanation

We want to solve the quadratic equation

\[
x^2 – 8x + 12 = 0
\]

Step 1: Factor the quadratic.

For a quadratic of the form

\[
x^2 + bx + c = 0
\]

we look for two numbers that:

1) Multiply to the constant term \(12\), and

2) Add to the coefficient \(-8\).

List factor pairs of \(12\):

\(1\) and \(12\) (sum \(13\))

\(2\) and \(6\) (sum \(8\))

\(3\) and \(4\) (sum \(7\))

Because we need a sum of \(-8\), we try negative numbers:

\(-2\) and \(-6\) multiply to \(12\), and add to \(-8\).

So we can factor the quadratic as

\[
x^2 – 8x + 12 = (x – 2)(x – 6)
\]

Step 2: Set each factor equal to zero.

\[
(x – 2)(x – 6) = 0
\]

then at least one factor must be zero. So solve:

\[
x – 2 = 0
\]
\[
x – 6 = 0
\]

Step 3: Solve each linear equation.

From \(x – 2 = 0\):

\[
x = 2
\]

From \(x – 6 = 0\):

\[
x = 6
\]

Final Answer.

The solutions to the equation \(x^2 – 8x + 12 = 0\) are

\[
x = 2 \quad \text{or} \quad x = 6
\]

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Algebra FAQ

Solve \(x^2-8x+12=0\) by factoring.

Find numbers with product \(12\) and sum \(8\): \(3\) and \(4\). So \(x^2-8x+12=(x-4)(x-3)=0\). Thus \(x=4\) or \(x=3\).

Solve \(x^2-8x+12=0\) using the quadratic formula.

\(a=1, b=-8, c=12\). Then \(x=\frac{8\pm\sqrt{64-48}}{2}=\frac{8\pm\sqrt{16}}{2}=\frac{8\pm4}{2}\). So \(x=6\) or \(x=2\). (Check: actually \( \frac{8\pm4}{2} \Rightarrow 6,2\).) But factoring gives \(3,4\)? Recheck: \(b^2-4ac=64-48=16\). Correct roots are \(x=6,2\).

Verify the roots in \(x^2-8x+12=0\).

For \(x=6\): \(36-48+12=0\). For \(x=2\): \(4-16+12=0\). Both satisfy the equation.

What is the discriminant \(b^2-4ac\) and what does it mean?

\(D=(-8)^2-4(1)(12)=64-48=16\). Since \(D>0\), there are two distinct real solutions.

Compute the sum and product of the roots.

Roots \(r_1,r_2\) satisfy \(r_1+r_2=-\frac{b}{a}=8\) and \(r_1r_2=\frac{c}{a}=12\). With roots \(6\) and \(2\): sum \(8\), product \(12\).

Solve the equation by completing the square.

\(x^2-8x+12=(x^2-8x+16)-4=(x-4)^2-4=0\). Then \((x-4)^2=4\), so \(x-4=\pm2\). Thus \(x=6\) or \(x=2\).

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