Solution Sets of Quadratic Equations Practice

Question

Complete this assessment to review what you’ve learned. It will not count toward your grade.
3 of 53 of 5 Items

Question
Use the table to answer the question.

x Substituted Evaluate True Statement?
7−11−−√

7
11−7–√

7+11−−√

11+7–√

For the quadratic equation (x−7)2=11
, complete the table by testing each of the given values to determine whether it is a solution. Identify the two solutions to the quadratic equation.

(1 point)
The smaller of the two solutions is x=
. The larger of the two solutions is x=
.
Remaining Attempts : 3
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Answer 1

To solve the quadratic equation \((x - 7)^2 = 11\), we'll first find the solutions by isolating \(x\).

1. Take the square root of both sides: \[ (x - 7) = \pm \sqrt{11} \]

2. Solve for \(x\):

   • For the positive case: \[ x - 7 = \sqrt{11} \implies x = 7 + \sqrt{11} \]

   • For the negative case: \[ x - 7 = -\sqrt{11} \implies x = 7 - \sqrt{11} \]

So the two solutions to the equation are:

• \(x_1 = 7 - \sqrt{11}\) (the smaller solution)
• \(x_2 = 7 + \sqrt{11}\) (the larger solution)

Now, let's fill in the table with the necessary evaluations.

• Substituting \(7 - \sqrt{11}\): \[ (7 - \sqrt{11}) - 7 = -\sqrt{11} \] Squaring this gives: \[ (-\sqrt{11})^2 = 11 \quad \text{(True statement)} \]

• Substituting \(7\): \[ 7 - 7 = 0 \] Squaring this gives: \[ 0^2 = 0 \quad \text{(not equal to 11, False statement)} \]

• Substituting \(7 + \sqrt{11}\): \[ (7 + \sqrt{11}) - 7 = \sqrt{11} \] Squaring this gives: \[ (\sqrt{11})^2 = 11 \quad \text{(True statement)} \]

• Substituting \(11 - 7\) (which is \(4\)): \[ 4 - 7 = -3 \] Squaring this gives: \[ (-3)^2 = 9 \quad \text{(not equal to 11, False statement)} \]

In summary:

• The smaller of the two solutions is \(x = 7 - \sqrt{11}\).
• The larger of the two solutions is \(x = 7 + \sqrt{11}\).

Finally, the answers are:

• The smaller of the two solutions is \(x = 7 - \sqrt{11}\).
• The larger of the two solutions is \(x = 7 + \sqrt{11}\).