Lesson Topics Discussion Quiz: Class Homework |
Steps-5 |
Title: Calculus |
Grade Lesson s6-p1 |
Explanation: Hello Students, time to practice and review the steps for the problem. |
Quiz: Discussion Step
| Id | Type | Name | Note |
|---|---|---|---|
1 |
Problem |
Solve the equation: \$(x + 1)/(x − 3) - (3x − 2)/(x + 2) = 2\$. |
|
2 |
Step |
The given equation |
\$(x + 1)/(x − 3) − (3x − 2)/(x + 2) = 2\$ |
3 |
Step |
To solve this equation, let’s find the least common denominator (LCD) of the rational expressions, which is (x − 3) (x + 2): |
\$((x + 1)(x + 2) − (3x − 2)(x − 3))/((x − 3)(x + 2))\$ = 2 |
4 |
Step |
Expanding and simplifying the numerator: |
\$((x^2 + 3x + 2) − (3x^2 − 11x + 6))/((x − 3)(x + 2)) = 2\$ \$(−2x^2 + 14x − 4)/((x − 3)(x + 2)) = 2/1\$ |
5 |
Step |
By using cross multiplication, and then simplified |
\$- 2x^2 + 14x − 4 = 2(x − 3)(x + 2)\$ \$- 2x^2 + 14x − 4 = 2(x2 − x − 6)\$ \$- 2x^2 + 14x − 4 = 2x^2 - 2x - 12\$ |
6 |
Step |
Bringing all terms to one side,then simplified |
\$2x^2 - 2x - 12 + 2x^2 - 14x + 4 = 0\$ \$x^2 - 4x - 2 = 0\$ |
7 |
Formula |
Now, we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula is x = \$((−b) ± \sqrt(b^2 − 4ac))/(2a)\$. |
|
8 |
Step |
In this case, a = 1, b = - 4, and c = - 2. Plugging these values into the quadratic formula, we have: |
x = \$(−(−4) ± \sqrt((−4)^2 − 4⋅(1) (−2)))/(2⋅1)\$ x = \$(4 ± 2\sqrt(6))/2\$ x = \$2 ± \sqrt(6)\$ |
9 |
Solution |
Therefore, the solutions to the original equation are x = \$2 + \sqrt(6)\$ and x = \$2 − \sqrt(6)\$. |
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10 |
Sumup |
Please summarize Problem, Clue, Hint, Formula, Steps and Solution |
|
Choices |
|||
11 |
Choice-A |
This option is incorrect because solutions don’t match the ones we found, which were x = \$2 + \sqrt(6), 2 − \sqrt(6)\$ |
Wrong \$2 - \sqrt(6)\$, \$-2 - \sqrt(6)\$ |
12 |
Choice-B |
This option is incorrect solution, x = \$2 + \sqrt(6)\$, but pairs it with x = \$2 − \sqrt(6)\$ . which doesn’t match the solution derived from our calculation |
Wrong \$2 + \sqrt(6)\$, \$-2 - \sqrt(6)\$ |
13 |
Choice-C |
This option is incorrect because it only partially matches the correct set of solutions |
Wrong \$-2 + \sqrt(6)\$, \$2 - \sqrt(6)\$ |
14 |
Choice-D |
This option is correct it suggests both solutions involve 2 as their base value, but with different signs for the square root term |
Correct \$2 + \sqrt(6)\$, \$2 - \sqrt(6)\$ |
15 |
Answer |
Option |
D |
16 |
Sumup |
Please summarize choices |
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