Certainly! Let's solve the quadratic equation [tex]\( y = x^2 - 8x + 15 \)[/tex]. We're looking for the values of [tex]\( x \)[/tex] that make [tex]\( y \)[/tex] equal to zero.
1. Set the quadratic equation equal to zero:
[tex]\[
x^2 - 8x + 15 = 0
\][/tex]
2. To solve this quadratic equation, we can factor it. We'll look for two numbers that multiply to the constant term (15) and add up to the coefficient of the linear term (-8).
3. The numbers that work are -3 and -5, because:
[tex]\[
-3 \times -5 = 15 \quad \text{and} \quad -3 + (-5) = -8
\][/tex]
4. Rewrite the quadratic equation using these factors:
[tex]\[
(x - 3)(x - 5) = 0
\][/tex]
5. Apply the zero-product property, which states that if the product of two factors is zero, at least one of the factors must be zero. So we set each factor equal to zero:
[tex]\[
x - 3 = 0 \quad \text{or} \quad x - 5 = 0
\][/tex]
6. Solve each equation for [tex]\( x \)[/tex]:
[tex]\[
x - 3 = 0 \implies x = 3
\][/tex]
[tex]\[
x - 5 = 0 \implies x = 5
\][/tex]
Thus, the solutions to the quadratic equation [tex]\( x^2 - 8x + 15 = 0 \)[/tex] are [tex]\( x = 3 \)[/tex] and [tex]\( x = 5 \)[/tex].
