National 5 - Using the Discriminant worksheet - Free Printable

To solve the problem, we need to use the discriminant of a quadratic equation. The discriminant is given by the formula:

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

where \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation in the form \(ax^2 + bx + c = 0\).

The discriminant tells us the nature of the roots of the quadratic equation:
- If \(\Delta > 0\), the equation has 2 real distinct roots.
- If \(\Delta = 0\), the equation has 1 real, repeated root.
- If \(\Delta < 0\), the equation has no real roots (the roots are complex).

Let's analyze each quadratic equation step by step.

---

1. \(x^2 + 3x + 2\)

- Coefficients: \(a = 1\), \(b = 3\), \(c = 2\)
- Discriminant:
\[
\Delta = b^2 - 4ac = 3^2 - 4(1)(2) = 9 - 8 = 1
\]
- Since \(\Delta > 0\), the equation has 2 real distinct roots.

---

2. \(x^2 + x + 2\)

- Coefficients: \(a = 1\), \(b = 1\), \(c = 2\)
- Discriminant:
\[
\Delta = b^2 - 4ac = 1^2 - 4(1)(2) = 1 - 8 = -7
\]
- Since \(\Delta < 0\), the equation has no real roots.

---

3. \(x^2 + 10x + 25\)

- Coefficients: \(a = 1\), \(b = 10\), \(c = 25\)
- Discriminant:
\[
\Delta = b^2 - 4ac = 10^2 - 4(1)(25) = 100 - 100 = 0
\]
- Since \(\Delta = 0\), the equation has 1 real, repeated root.

---

4. \(x^2 - 3x + 2\)

- Coefficients: \(a = 1\), \(b = -3\), \(c = 2\)
- Discriminant:
\[
\Delta = b^2 - 4ac = (-3)^2 - 4(1)(2) = 9 - 8 = 1
\]
- Since \(\Delta > 0\), the equation has 2 real distinct roots.

---

5. \(x^2 - 4x - 5\)

- Coefficients: \(a = 1\), \(b = -4\), \(c = -5\)
- Discriminant:
\[
\Delta = b^2 - 4ac = (-4)^2 - 4(1)(-5) = 16 + 20 = 36
\]
- Since \(\Delta > 0\), the equation has 2 real distinct roots.

---

6. \(x^2 - 2x + 1\)

- Coefficients: \(a = 1\), \(b = -2\), \(c = 1\)
- Discriminant:
\[
\Delta = b^2 - 4ac = (-2)^2 - 4(1)(1) = 4 - 4 = 0
\]
- Since \(\Delta = 0\), the equation has 1 real, repeated root.

---

7. \(2x^2 + 5x + 2\)

- Coefficients: \(a = 2\), \(b = 5\), \(c = 2\)
- Discriminant:
\[
\Delta = b^2 - 4ac = 5^2 - 4(2)(2) = 25 - 16 = 9
\]
- Since \(\Delta > 0\), the equation has 2 real distinct roots.

---

8. \(3x^2 - 4x + 1\)

- Coefficients: \(a = 3\), \(b = -4\), \(c = 1\)
- Discriminant:
\[
\Delta = b^2 - 4ac = (-4)^2 - 4(3)(1) = 16 - 12 = 4
\]
- Since \(\Delta > 0\), the equation has 2 real distinct roots.

---

Final Answers:

\[
\begin{array}{|c|c|}
\hline
x^2 + 3x + 2 & \text{2 real distinct roots} \\
\hline
x^2 + x + 2 & \text{no real roots} \\
\hline
x^2 + 10x + 25 & \text{1 real, repeated root} \\
\hline
x^2 - 3x + 2 & \text{2 real distinct roots} \\
\hline
x^2 - 4x - 5 & \text{2 real distinct roots} \\
\hline
x^2 - 2x + 1 & \text{1 real, repeated root} \\
\hline
2x^2 + 5x + 2 & \text{2 real distinct roots} \\
\hline
3x^2 - 4x + 1 & \text{2 real distinct roots} \\
\hline
\end{array}
\]

\boxed{
\begin{array}{|c|c|}
\hline
x^2 + 3x + 2 & \text{2 real distinct roots} \\
\hline
x^2 + x + 2 & \text{no real roots} \\
\hline
x^2 + 10x + 25 & \text{1 real, repeated root} \\
\hline
x^2 - 3x + 2 & \text{2 real distinct roots} \\
\hline
x^2 - 4x - 5 & \text{2 real distinct roots} \\
\hline
x^2 - 2x + 1 & \text{1 real, repeated root} \\
\hline
2x^2 + 5x + 2 & \text{2 real distinct roots} \\
\hline
3x^2 - 4x + 1 & \text{2 real distinct roots} \\
\hline
\end{array}
}