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Determine whether the values -1 and 7/3 are solutions to the quadratic equation 3x2−4x−4=3


Responses

Both x=−1 and x=7/3 are solutions.

Only x=7/3 is a solution.

Neither x=−1 nor x=7/3 is a solution.

Only x=−1 is a solution.

5 answers

To determine whether the values \(-1\) and \(\frac{7}{3}\) are solutions to the equation \(3x^2 - 4x - 4 = 3\), we first simplify the equation. We can rewrite it as:

\[ 3x^2 - 4x - 4 - 3 = 0 \]

This simplifies to:

\[ 3x^2 - 4x - 7 = 0 \]

Now, we will substitute each candidate value into this simplified equation and check if it satisfies the equation.

Testing \(x = -1\):

Substituting \(x = -1\):

\[ 3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7 = 3 + 4 - 7 = 0 \]

Since the left-hand side equals 0, \(x = -1\) is a solution.

Testing \(x = \frac{7}{3}\):

Substituting \(x = \frac{7}{3}\):

\[ 3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7 \] Calculating \(\left(\frac{7}{3}\right)^2\):

\[ \left(\frac{7}{3}\right)^2 = \frac{49}{9} \]

Then,

\[ 3\left(\frac{49}{9}\right) - 4\left(\frac{7}{3}\right) - 7 = \frac{147}{9} - \frac{28}{3} - 7 \] Converting \(\frac{28}{3}\) and \(7\) to ninths:

\[ \frac{28}{3} = \frac{84}{9} \quad \text{and} \quad 7 = \frac{63}{9} \]

Now substituting back:

\[ \frac{147}{9} - \frac{84}{9} - \frac{63}{9} = \frac{147 - 84 - 63}{9} = \frac{0}{9} = 0 \]

Since the left-hand side equals 0, \(x = \frac{7}{3}\) is also a solution.

Conclusion:

Both \(x = -1\) and \(x = \frac{7}{3}\) are solutions.

Therefore, the response is:

Both \(x = -1\) and \(x = \frac{7}{3}\) are solutions.

Show how many solutions are in the solution set for the equation 3x^2=21

Responses:

infinitely many

two

one

zero

To solve the equation \(3x^2 = 21\), we can start by isolating \(x^2\). Here are the steps:

  1. Divide both sides of the equation by 3:

    \[ x^2 = \frac{21}{3} = 7 \]

  2. Now we take the square root of both sides:

    \[ x = \pm \sqrt{7} \]

This gives us two solutions:

\[ x = \sqrt{7} \quad \text{and} \quad x = -\sqrt{7} \]

Conclusion:

The equation \(3x^2 = 21\) has two distinct solutions in the solution set.

Therefore, the correct response is:

two.

Show how many solutions are in the solution set for 2(x−5)^2+2=20

Responses:

zero

infinitely many

one

two

To find the number of solutions for the equation \(2(x-5)^2 + 2 = 20\), we will first simplify it step by step.

  1. Start by isolating the squared term:

    \[ 2(x-5)^2 + 2 = 20 \]

    Subtract 2 from both sides:

    \[ 2(x-5)^2 = 18 \]

  2. Divide both sides by 2:

    \[ (x-5)^2 = 9 \]

  3. Now take the square root of both sides:

    \[ x - 5 = \pm 3 \]

    This gives us two equations to solve:

    • \(x - 5 = 3\)
    • \(x - 5 = -3\)

    Solving these equations:

    For the first equation:

    \[ x - 5 = 3 \implies x = 8 \]

    For the second equation:

    \[ x - 5 = -3 \implies x = 2 \]

Thus, the solutions are \(x = 8\) and \(x = 2\).

Conclusion:

The equation \(2(x - 5)^2 + 2 = 20\) has two distinct solutions in the solution set.

Therefore, the correct response is:

two.