Q. \(x^2 – 8x + 15 = 0\)

Q: How do I factor x^2-8x+15=0?

Find numbers with product 15 and sum -8: -3 and -5. So (x-3)(x-5)=0.

Q: What are the roots of x^2-8x+15=0?

Using (x-3)(x-5)=0, set each factor to zero: x-3=0 gives x=3, and x-5=0 gives x=5.

Q: Can I solve it using the quadratic formula?

For x^2-8x+15=0, a=1, b=-8, c=15. Then x=\frac{8\pm\sqrt{64-60}}{2}=\frac{8\pm 2}{2}, so x=3 or x=5.

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

Discriminant \Delta=b^2-4ac=64-60=4. Since \Delta>0, there are two distinct real roots.

Q: How do I check my solutions x=3 and x=5?

Substitute: 3^2-8(3)+15=9-24+15=0. Also 5^2-8(5)+15=25-40+15=0.

Q: What is the vertex of the parabola y=x^2-8x+15?

Vertex x-coordinate is -\frac{b}{2a}=\frac{8}{2}=4. The parabola is y=(x-4)^2-1, so the minimum value is -1.

Answer

We solve the quadratic \(x^2-8x+15=0\) by factoring.

Look for two numbers that multiply to \(15\) and add to \(-8\). They are \(-3\) and \(-5\).

\[
x^2-8x+15=(x-3-5)(x-3? ) \text{ (factorization check)}
\]

Correct factorization:

\[
x^2-8x+15=(x-3)(x-5)=0
\]

So \(x-3=0\) or \(x-5=0\).

\[
x=3,\quad x=5
\]

Final result: \(x=3\) or \(x=5\).

Detailed Explanation

We want to solve the quadratic equation

\[
x^2 – 8x + 15 = 0
\]

Step 1: Factor the quadratic.

For a quadratic in the form \(ax^2 + bx + c\), we look for two numbers that multiply to \(ac\) and add to \(b\).

Here, \(a = 1\), \(b = -8\), and \(c = 15\).

\[
ac = 1 \cdot 15 = 15
\]

We need two numbers whose product is \(15\) and whose sum is \(-8\).

Numbers that work are \(-3\) and \(-5\) because:

\[
(-3)(-5) = 15
\]

\[
-3 + (-5) = -8
\]

So we can factor the quadratic as:

\[
x^2 – 8x + 15 = (x – 3)(x – 5)
\]

Step 2: Set each factor equal to zero.

\[
(x – 3)(x – 5) = 0
\]

then at least one factor must be zero.

\[
x – 3 = 0
\]

\[
x – 5 = 0
\]

Step 3: Solve each simple equation.

From \(x – 3 = 0\):

\[
x = 3
\]

From \(x – 5 = 0\):

\[
x = 5
\]

Final Answer:

\[
x = 3 \text{ or } x = 5
\]

See full solution

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Algebra FAQ

How do I factor \(x^2-8x+15=0\)?

Find numbers with product \(15\) and sum \(-8\): \(-3\) and \(-5\). So \((x-3)(x-5)=0\).

What are the roots of \(x^2-8x+15=0\)?

Using \((x-3)(x-5)=0\), set each factor to zero: \(x-3=0\) gives \(x=3\), and \(x-5=0\) gives \(x=5\).

Can I solve it using the quadratic formula?

For \(x^2-8x+15=0\), \(a=1\), \(b=-8\), \(c=15\). Then \(x=\frac{8\pm\sqrt{64-60}}{2}=\frac{8\pm 2}{2}\), so \(x=3\) or \(x=5\).

What is the discriminant, and what does it mean?

Discriminant \( \Delta=b^2-4ac=64-60=4\). Since \(\Delta>0\), there are two distinct real roots.

How do I check my solutions \(x=3\) and \(x=5\)?

Substitute: \(3^2-8(3)+15=9-24+15=0\). Also \(5^2-8(5)+15=25-40+15=0\).

What is the vertex of the parabola \(y=x^2-8x+15\)?

Vertex \(x\)-coordinate is \(-\frac{b}{2a}=\frac{8}{2}=4\). The parabola is \(y=(x-4)^2-1\), so the minimum value is \(-1\).

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