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or the quadratic equation −2x2−3x+2=0

, complete the table by testing each of the given values to determine whether it is a solution. Identify which one of the values is in the solution set. Only x =

is in the solution set.
Question 2
2. Solve x2=289
by inspection. There are two real solutions. Enter the lesser number first.(2 points)
{
,
}
Question 3
3. Which values are in the solution set of the quadratic equation, −9+(x−3)2−4=−4
? Solve using square roots and choose all that apply.(2 points)
Responses

-8
-8

-22
-22

8
8

2
2

16
16

-2
-2

28
28

-10
-10
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1 answer

Let's tackle each question step by step.

Question 1

Given Quadratic Equation: \[ -2x^2 - 3x + 2 = 0\]

We will test the given values (you didn't specify which values to test, so you will need to input the values you're given). Let's assume we test with \( x = -2, -1, 0, 1, 2 \).

  1. If \( x = -2 \): \[ -2(-2)^2 - 3(-2) + 2 = -2(4) + 6 + 2 = -8 + 6 + 2 = 0
    \text{ (solution)} \]

  2. If \( x = -1 \): \[ -2(-1)^2 - 3(-1) + 2 = -2(1) + 3 + 2 = -2 + 3 + 2 = 3
    \text{ (not a solution)} \]

  3. If \( x = 0 \): \[ -2(0)^2 - 3(0) + 2 = 0 + 0 + 2 = 2
    \text{ (not a solution)} \]

  4. If \( x = 1 \): \[ -2(1)^2 - 3(1) + 2 = -2 - 3 + 2 = -3
    \text{ (not a solution)} \]

  5. If \( x = 2 \): \[ -2(2)^2 - 3(2) + 2 = -2(4) - 6 + 2 = -8 - 6 + 2 = -12
    \text{ (not a solution)} \]

The solution set from this example:

  • Only \( x = -2 \) is in the solution set.

Question 2

Solve: \[ x^2 = 289\]

Taking the square root on both sides: \[ x = \sqrt{289} \quad \text{or} \quad x = -\sqrt{289} \] Thus, the solutions are: \[ x = 17 \quad \text{and} \quad x = -17 \]

The lesser number first: \[ {-17, 17} \]

Question 3

Given Quadratic Equation: \[ -9 + (x - 3)^2 - 4 = -4\]

First, simplify the equation: \[ -9 + (x - 3)^2 - 4 + 4 = 0 \] So we have: \[ -9 + (x - 3)^2 = 0 \] This simplifies to: \[ (x - 3)^2 = 9 \]

Now, take the square root of both sides: \[ x - 3 = 3 \quad \text{or} \quad x - 3 = -3 \] Solving these gives:

  1. \(x - 3 = 3 \) ⇒ \( x = 6\)
  2. \( x - 3 = -3 \) ⇒ \( x = 0\)

The values that are in the solution set from given options are:

  • 8 (not in the set)
  • 2 (not in the set)
  • 16 (not in the set)
  • -2 (not in the set)
  • 28 (not in the set)
  • -10 (not in the set)

So the values 0 and 6 would be in the solution set if included.

Final Responses:

  • For Question 1: Only \( x = -2 \) is in the solution set.
  • For Question 2: \( {-17, 17} \).
  • For Question 3: \( x = 0 \) and \( x = 6\).