For the quadratic equation βˆ’2x2βˆ’3x+2=0

βˆ’
2
π‘₯
2
βˆ’
3
π‘₯
+
2
=
0
, determine if -2 and 2 are solutions.

(1 point)
Responses

Both x=βˆ’2
π‘₯
=
βˆ’
2
and x=2
π‘₯
=
2
are solutions.
Both x is equal to negative 2 and x is equal to 2 are solutions.

Only x=βˆ’2
π‘₯
=
βˆ’
2
is a solution.
Only x is equal to negative 2 is a solution.

Neither x=βˆ’2
π‘₯
=
βˆ’
2
nor x=2
π‘₯
=
2
is a solution.
Neither x is equal to negative 2 nor x is equal to 2 is a solution.

Only x=2
π‘₯
=
2
is a solution.

9 answers

To determine if \( x = -2 \) and \( x = 2 \) are solutions to the quadratic equation \( -2x^2 - 3x + 2 = 0 \), we can substitute these values into the equation and see if they satisfy it.

  1. Testing \( x = -2 \): \[ -2(-2)^2 - 3(-2) + 2 = -2(4) + 6 + 2 = -8 + 6 + 2 = 0 \] Since the result is \( 0 \), \( x = -2 \) is a solution.

  2. Testing \( x = 2 \): \[ -2(2)^2 - 3(2) + 2 = -2(4) - 6 + 2 = -8 - 6 + 2 = -12 \] Since the result is \( -12 \), \( x = 2 \) is not a solution.

Based on this evaluation, the correct response is: Only \( x = -2 \) is a solution.

Consider the equation x2=0
π‘₯
2
=
0
. How many solutions are there?(1 point)
Responses

None
None

One
One

Two
Two

Infinitely Many

The equation \( x^2 = 0 \) has one solution. This is because the only value of \( x \) that satisfies the equation is \( x = 0 \).

Thus, the correct response is: One.

Solve x2=78

by inspection. There are two real solutions. Enter the lesser number first and use exponents in your responses.

To solve the equation \( x^2 = 78 \) by inspection, we can determine the possible values of \( x \).

The solutions are: \[ x = \sqrt{78} \quad \text{and} \quad x = -\sqrt{78} \]

Calculating \( \sqrt{78} \): \[ \sqrt{78} \approx 8.83 \]

So the two solutions are: \[ -\sqrt{78} \quad \text{and} \quad \sqrt{78} \]

When entering the solutions, with the lesser number first, the response would be: \[ -\sqrt{78}, \sqrt{78} \]

Solve the following quadratic equation using square roots: βˆ’9+(xβˆ’3)^2/βˆ’4=βˆ’4

.(1 point)
Responses

{16, βˆ’10}
{
16
,

βˆ’
10
}
{16, βˆ’10}
{
16
,

βˆ’
10
}

{βˆ’2, 8}
{
βˆ’
2
,

8
}
{βˆ’2, 8}
{
βˆ’
2
,

8
}

{28, βˆ’22}

To solve the equation

\[ -9 + \frac{(x - 3)^2}{-4} = -4, \]

let's first isolate the term with \( (x - 3)^2 \).

  1. Start by moving -9 to the other side: \[ \frac{(x - 3)^2}{-4} = -4 + 9 \] \[ \frac{(x - 3)^2}{-4} = 5 \]

  2. Multiply both sides by -4: \[ (x - 3)^2 = -20 \]

  3. Since \( (x - 3)^2 \) equals a negative number (-20), this indicates that there are no real solutions. The square of a real number cannot be negative.

Thus, the quadratic equation has no real solutions.

However, if you're looking for potential answers from the options provided, none of them are correct since you have no valid solutions. Therefore, none of the sets {16, -10}, {-2, 8}, or {28, -22} apply.

If complex solutions were to be considered:

  1. Taking square roots would yield: \[ x - 3 = \pm i \sqrt{20} \] \[ x - 3 = \pm 2i \sqrt{5} \] \[ x = 3 \pm 2i \sqrt{5} \]

Thus the complex solutions would be \( 3 + 2i \sqrt{5} \) and \( 3 - 2i \sqrt{5} \), but since the question seems to based on real solutions, the answer is that there are no real solutions.

Solve the following quadratic equation using square roots: βˆ’9+(xβˆ’3)^2/βˆ’4=βˆ’4

.(1 point)
Responses

{16, βˆ’10}

{16, βˆ’10}

{βˆ’2, 8}

{βˆ’2, 8-}{28, βˆ’22}

Let's solve the equation

\[ -9 + \frac{(x - 3)^2}{-4} = -4. \]

  1. First, isolate the term with \( (x - 3)^2 \):

    \[ -9 + \frac{(x - 3)^2}{-4} = -4 \]

    Add 9 to both sides:

    \[ \frac{(x - 3)^2}{-4} = 5 \]

  2. Multiply both sides by -4 to remove the fraction:

    \[ (x - 3)^2 = -20 \]

  3. Since \( (x - 3)^2 = -20 \) has no real solutions (a square cannot be negative in real numbers), let's see if we can simplify our work:

  4. Let's rewrite that:

    \[ (x - 3)^2 = 20i^2 \quad \text{(where } i = \sqrt{-1}\text{)} \]

  5. Taking the square root gives us:

    \[ x - 3 = \pm \sqrt{20}i \]

    This further simplifies to:

    \[ x - 3 = \pm 2\sqrt{5}i \]

  6. Thus, the complex solutions are:

    \[ x = 3 \pm 2\sqrt{5}i \]

Since the equation has no real solutions, the potential responses you provided (which suggest real outputs) cannot apply here.

However, it appears there was a misunderstanding in how to rework the equation in earlier messages. Let's double-check if the intent was to look for real solutions from already provided options or some other misunderstanding might have occurred.

Since none of the responses {16, -10}, {-2, 8}, {28, -22} represent valid real solutions, the equation \( -9 + \frac{(x - 3)^2}{-4} = -4 \) actually doesn't yield real solutions. If you're facing template options or misunderstandings based on incorrect values formulated, lets recheck the question context.

Final Answer

In conclusion, the provided options seem misaligned with the equation's resolution. The equation yields complex solutions, and none of the answers match itβ€”no real solutions are valid.