Question

Complete the square to identify the two values that solve the following quadratic equation: x2+8x=9
.(1 point)
x=

2. Using the quadratic formula, find the solution to x2+6x+9=0
3.Solve −2x2+3x−1=−2
using the quadratic formula. Round the solution(s) to the nearest hundredths. Enter the smaller solution first.(1 point)
x=
or
4.
.
How many real solutions does the following quadratic equation have?

25x2+60x+36=0

(1 point)
solution(s)

Answer 1

Let's solve each question step by step:

1. **Completing the square for the equation \(x^2 + 8x = 9\):**

To complete the square, follow these steps:

- Move the constant to the other side:
\[
x^2 + 8x - 9 = 0
\]

- Take half of the coefficient of \(x\) (which is \(8\)), square it, and add it to both sides:
\[
\left(\frac{8}{2}\right)^2 = 16
\]
Adding \(16\) to both sides gives:
\[
x^2 + 8x + 16 = 9 + 16
\]
- This simplifies to:
\[
(x + 4)^2 = 25
\]

- Now take the square root of both sides:
\[
x + 4 = \pm 5
\]

- Solve for \(x\):
\[
x + 4 = 5 \quad \Rightarrow \quad x = 1
\]
\[
x + 4 = -5 \quad \Rightarrow \quad x = -9
\]

So the solutions are:
\[
x = 1 \quad \text{or} \quad x = -9
\]

2. **Using the quadratic formula for \(x^2 + 6x + 9 = 0\):**

The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 1\), \(b = 6\), and \(c = 9\).

- Calculate the discriminant:
\[
b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0
\]

- Since the discriminant is \(0\), there is one real solution:
\[
x = \frac{-6 \pm 0}{2 \cdot 1} = \frac{-6}{2} = -3
\]

Thus, the solution is:
\[
x = -3
\]

3. **Solve \(-2x^2 + 3x - 1 = -2\) using the quadratic formula:**

First, rearrange the equation:
\[
-2x^2 + 3x - 1 + 2 = 0 \quad \Rightarrow \quad -2x^2 + 3x + 1 = 0
\]

Multiply through by \(-1\) to make \(a\) positive:
\[
2x^2 - 3x - 1 = 0
\]

Here, \(a = 2\), \(b = -3\), and \(c = -1\).

- Calculate the discriminant:
\[
b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot (-1) = 9 + 8 = 17
\]

- Using the quadratic formula:
\[
x = \frac{-(-3) \pm \sqrt{17}}{2 \cdot 2} = \frac{3 \pm \sqrt{17}}{4}
\]

- We can find the two solutions:
\[
x_1 = \frac{3 + \sqrt{17}}{4} \quad \text{(larger)}
\]
\[
x_2 = \frac{3 - \sqrt{17}}{4} \quad \text{(smaller)}
\]

- Calculate the approximations:
\[
\sqrt{17} \approx 4.123
\]
\[
x_2 \approx \frac{3 - 4.123}{4} \approx \frac{-1.123}{4} \approx -0.2808 \quad \text{(rounds to -0.28)}
\]
\[
x_1 \approx \frac{3 + 4.123}{4} \approx \frac{7.123}{4} \approx 1.78075 \quad \text{(rounds to 1.78)}
\]

Thus the solutions are:
\[
x = -0.28 \quad \text{or} \quad x = 1.78
\]

4. **Number of real solutions for the equation \(25x^2 + 60x + 36 = 0\):**

Use the discriminant to determine the number of solutions:
\[
b^2 - 4ac = 60^2 - 4 \cdot 25 \cdot 36 = 3600 - 3600 = 0
\]

Since the discriminant is \(0\), there is **exactly one real solution**.

Thus, summarizing the answers:
1. \(x = 1\) or \(x = -9\)
2. \(x = -3\)
3. \(x = -0.28\) or \(x = 1.78\)
4. The number of solution(s): \(1\) real solution.