Question

Jackie correctly solved this equation: 4 x − 2 1 x 2 − 4 = 1 x 2 . which statement describes the solutions she found?

a. she found one valid solution and no extraneous solutions.
b. she found two valid solutions and no extraneous solutions.
c. she found one valid solution and one extraneous solution.
d. she found no valid solutions and two extraneous solutions.

Answer 1

Jackie correctly solved the quadratic equation by obtaining two valid solutions and no extraneous solutions.

Step-by-step explanation:

Jackie correctly solved the equation 4x − 21x^2 − 4 = 1x^2. To determine the number of solutions she found, we need to analyze the equation. Since it is a quadratic equation, it can have at most two solutions. We can solve the equation to find the actual solutions. By rearranging the equation and using the quadratic formula, we find that it simplifies to x^2 + 1.2x - 6 = 0. When we apply the quadratic formula, we obtain two valid solutions: x = -3 and x = 2.

Therefore, the correct statement describing the solutions Jackie found is option b: she found two valid solutions and no extraneous solutions.

Answer 2

we can conclude that Jackie found two valid solutions and no extraneous solutions. The answer is:

b. she found two valid solutions and no extraneous solutions.

To analyze the solutions that Jackie may have found, we need to solve the equation step by step:

`\[ 4x - (2)/(1x^2 - 4) = (1)/(x^2) \]`

This equation appears to have rational expressions, and it's common when solving rational equations to find extraneous solutions—solutions that don't satisfy the original equation. Let's solve it:

First, note that the equation has denominators that can be factored:

`\[ 1x^2 - 4 = (x + 2)(x - 2) \]`

The common denominator for the entire equation will be
`\( x^2(x + 2)(x - 2) \)`. We'll multiply each term by this common denominator to clear the fractions:

`\[ 4x(x^2(x + 2)(x - 2)) - 2(x^2) = (x + 2)(x - 2) \]`

`\[ 4x^4 - 2x^2 = 1 \]`

Now we have a polynomial equation. Let's rearrange it and solve for
`\( x^2 \)`:

`\[ 4x^4 - 2x^2 - 1 = 0 \]`

This is a quadratic in form equation, and we can use a substitution
`\( u = x^2 \)` to make it a standard quadratic equation:

`\[ 4u^2 - 2u - 1 = 0 \]`

Solving for
`\( u \)` using the quadratic formula:

`\[ u = (-(-2) \pm √((-2)^2 - 4 \cdot 4 \cdot (-1)))/(2 \cdot 4) \]`

`\[ u = (2 \pm √(4 + 16))/(8) \]`

`\[ u = (2 \pm √(20))/(8) \]`

`\[ u = (2 \pm 2√(5))/(8) \]`

`\[ u = (1 \pm √(5))/(4) \]`

So we have two solutions for
`\( u \)`:

`\[ u_1 = (1 + √(5))/(4) \]`

`\[ u_2 = (1 - √(5))/(4) \]`

Since
`\( u = x^2 \)`, we can now find
`\( x \)`:

`\[ x^2 = (1 + √(5))/(4) \]`

`\[ x = \pm \sqrt{(1 + √(5))/(4)} \]`

And

`\[ x^2 = (1 - √(5))/(4) \]`

`\[ x = \pm \sqrt{(1 - √(5))/(4)} \]`

But the second set of solutions for
`\( x \)` is not valid because the term inside the square root becomes negative (since
`\( √(5) > 1 \)`), which cannot produce a real number solution for
`\( x \)`.

So, we have two potential solutions from
`\( u_1 \)`, and we need to check if they are valid by substituting them back into the original equation to see if they don't make any denominators zero or otherwise invalidate the solution.

Let's check the potential solutions:

For
`\( x = \sqrt{(1 + √(5))/(4)} \)`, we need to ensure that this value of
`\( x \)` does not make the original denominators equal to zero:

`\[ 1x^2 - 4 = 0 \]`

`\[ (\sqrt{(1 + √(5))/(4)})^2 - 4 \\eq 0 \]`

`\[ (1 + √(5))/(4) - 4 \\eq 0 \]`

`\[ (1 + √(5) - 16)/(4) \\eq 0 \]`

`\[ 1 + √(5) - 16 \\eq 0 \]`

`\[ √(5) - 15 \\eq 0 \]`

This is true, so
`\( x = \sqrt{(1 + √(5))/(4)} \)` is a valid solution. The negative counterpart will also be valid since the denominators are squared terms.

Therefore, we can conclude that Jackie found two valid solutions and no extraneous solutions. The answer is:

b. she found two valid solutions and no extraneous solutions.