Question

Find the real solutions, if any, of each equation. Use the quadratic formula. $$ 9 t^{2}-6 t+1=0 $$

Short answer

The only solution is \(t = \frac{1}{3}\).

Step-by-step solution

- Identify coefficients

In the quadratic equation \(ax^2 + bx + c = 0\), identify the coefficients \(a\), \(b\), and \(c\). For the equation \(9t^2 - 6t + 1 = 0\): \(a = 9\), \(b = -6\), and \(c = 1\).

- Write the quadratic formula

The quadratic formula is given by \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].

- Substitute the coefficients

Substitute the identified coefficients \(a = 9\), \(b = -6\), and \(c = 1\) into the quadratic formula: \[t = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9}\].

- Simplify under the square root

Calculate the value under the square root (the discriminant): \[-6^2 - 4 \cdot 9 \cdot 1 = 36 - 36 = 0\].

- Solve using the quadratic formula

Since the discriminant is zero, there is only one solution: \[t = \frac{6}{18} = \frac{1}{3}\].

Key concepts

discriminant

When solving a quadratic equation, one of the key components to look at is the discriminant. The discriminant is part of the quadratic formula and can tell us important information about the nature of the solutions. It is located under the square root symbol in the quadratic formula, \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Specifically, the discriminant is the expression \(b^2 - 4ac\).

The value of the discriminant determines the number and type of solutions of the quadratic equation:

• If the discriminant is greater than zero, the quadratic equation has two distinct real solutions.
• If the discriminant is equal to zero, the quadratic equation has exactly one real solution, also known as a repeated or double root.
• If the discriminant is less than zero, the quadratic equation has no real solutions, but two complex solutions.

In our example, the discriminant calculation was \( (-6)^2 - 4 \cdot 9 \cdot 1 = 36 - 36 = 0 \), which means there is exactly one real solution.

quadratic equation

A quadratic equation is a second-degree polynomial equation in the form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants, and \(a\) is not equal to zero. The solutions to this equation are the values of \(x \) that make the equation true.

The quadratic formula is a general solution to any quadratic equation and is given by: \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula helps in finding the values of \(x\) by substituting the coefficients \(a, b, \) and \(c\).

In our exercise, the equation was \(9t^2 - 6t + 1 = 0 \). By comparing this to the general form \(ax^2 + bx + c = 0\), we identified:

• \(a = 9\)
• \(b = -6 \)
• \(c = 1\)

real solutions

Real solutions to a quadratic equation are the values of the variable that satisfy the equation and are real numbers. The quadratic formula, \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), can help find these solutions.

In our example, after calculating the discriminant, we found that the discriminant is zero. This tells us there is exactly one real solution. Substituting the values of \(a = 9\), \(b = -6\), and \(c = 1\) into the formula, we get:
<\(\frac{6 \pm \sqrt{0}}{18} = \frac{6}{18} = \frac{1}{3}\)

Therefore, the only real solution to the equation \(9t^2 - 6t + 1 = 0\) is \(\frac{1}{3}\).

To summarize, the real solutions of a quadratic equation depend on the value of the discriminant. If it is zero, there is one real solution, just like in this exercise.