Find the real​ solutions, if​ any, of the following equation. Use the quadratic formula. 49t^2 - 14t + 1 = 0

We are given the quadratic equation:

\[
49t^2 - 14t + 1 = 0
\]

We will solve this using the quadratic formula, which is given by:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \(49t^2 - 14t + 1 = 0\), the coefficients are:

- \(a = 49\)
- \(b = -14\)
- \(c = 1\)

### Step 1: Calculate the discriminant (\(\Delta\)).

\[
\Delta = b^2 - 4ac
\]

Substitute the values of \(a\), \(b\), and \(c\):

\[
\Delta = (-14)^2 - 4(49)(1) = 196 - 196 = 0
\]

### Step 2: Apply the quadratic formula.

Since \(\Delta = 0\), we have a single repeated real solution. Using the quadratic formula:

\[
t = \frac{-(-14) \pm \sqrt{0}}{2(49)} = \frac{14}{98} = \frac{1}{7}
\]

### Final Answer:
The solution is:

\[
t = \frac{1}{7}
\]

There is only one real solution since the discriminant is zero.

Would you like more details or have any other questions?

Here are 5 questions to expand on the topic:
1. How does the discriminant affect the number of real solutions in a quadratic equation?
2. What happens when the discriminant is negative?
3. Can you graph a quadratic equation to visualize its roots?
4. How can we factor quadratic equations instead of using the quadratic formula?
5. What are some real-life applications of quadratic equations?

**Tip:** When solving quadratic equations, always check the discriminant first—it helps you know how many real solutions to expect!

Explore the step-by-step solution to the quadratic equation 49t^2 - 14t + 1 using the quadratic formula. Learn how to apply the discriminant to determine the number of real solutions. Perfect for students in Grades 8-10 learning algebra and quadratic equations.

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