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  2. Topic
    Mathematics
  3. Study Set
    Intermediate Algebra
  4. Exam
    Exam 6: Factoring
  5. Question
    Find All Solutions by Factoring\(2 k^{2}=-9 \mathrm{k}-9\) A)\(\left\{3,-\frac{3}{2}\right\}\) B)\(\left\{3, \frac{3}{2}\right\}\)
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Find All Solutions by Factoring 2k2=−9k−92 k^{2}=-9 \mathrm{k}-92k2=−9k−9 A) {3,−32}\left\{3,-\frac{3}{2}\right\}{3,−23​} B) {3,32}\left\{3, \frac{3}{2}\right\}{3,23​}

Question 2

Question 2

Multiple Choice

Find all solutions by factoring.
- 2k2=−9k−92 k^{2}=-9 \mathrm{k}-92k2=−9k−9


A) {3,−32}\left\{3,-\frac{3}{2}\right\}{3,−23​}

B) {3,32}\left\{3, \frac{3}{2}\right\}{3,23​}

C) {−32,−3}\left\{-\frac{3}{2},-3\right\}{−23​,−3}

D) {−3,−6}\{-3,-6\}{−3,−6}

Correct Answer:

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