Solving the Quadratic Equation: t² – 6t + 8 = 0 — A Complete Guide

Understanding how to solve quadratic equations is a fundamental skill in algebra, essential for students, mathematicians, and professionals in various STEM fields. One classic example is the equation t² – 6t + 8 = 0, a beautifully simple quadratic with real-world applications. In this SEO-optimized article, we will explore how to solve this equation step-by-step, understand its meaning, and discover its practical uses.

Understanding the Context

What is t² – 6t + 8 = 0?

The equation t² – 6t + 8 = 0 is a quadratic equation in standard form:
at² + bt + c = 0, where a = 1, b = –6, and c = 8.

Quadratic equations describe parabolic relationships and are crucial in physics, engineering, economics, and many other disciplines. The solutions (roots) of this equation tell us the values of t where the quadratic function f(t) = t² – 6t + 8 equals zero — i.e., the points where the parabola intersects the t-axis.

t^2 - 6t + 8 = 0

Key Insights

Step-by-Step Solution to t² – 6t + 8 = 0

Method 1: Factoring

Factoring is often the fastest way when the quadratic expression can be broken down into simpler binomials.

  1. Identify two numbers that multiply to c = 8 and add to b = –6.
    These numbers are –2 and –4, since:
    (-2) × (-4) = 8
    (-2) + (–4) = –6

  2. Write the factored form:
    t² – 6t + 8 = (t – 2)(t – 4)

  3. Apply the Zero Product Property:
    If (t – 2)(t – 4) = 0, then either:
    t – 2 = 0 → t = 2
    or t – 4 = 0 → t = 4

Continue Reading

A tank is filled by two pipes. The first pipe alone fills it in 6 hours, and the second pipe in 4 hours. How long will it take to fill the tank if both pipes are opened together?
The rate of the first pipe is \(\frac{1}{6}\) of the tank per hour.
The rate of the second pipe is \(\frac{1}{4}\) of the tank per hour.

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Final Thoughts

✅ Solutions:
t = 2 and t = 4

Method 2: The Quadratic Formula

For any quadratic at² + bt + c = 0, the solutions are:
t = [–b ± √(b² – 4ac)] / (2a)

Plug in a = 1, b = –6, c = 8:

  1. Compute discriminant:
    Δ = b² – 4ac = (–6)² – 4(1)(8) = 36 – 32 = 4

  2. Take square root of Δ:
    √Δ = √4 = 2

  1. Plug into formula:
    t = [–(–6) ± 2] / (2×1) = [6 ± 2] / 2

  2. Calculate both solutions:
    t = (6 + 2)/2 = 8/2 = 4
    t = (6 – 2)/2 = 4/2 = 2

Result: t = 2 and t = 4 — same as before.