The solution set is: q = {2, -2) ; [ or, write as: q = {± 2} ] .
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Explanation:
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GIven:
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7q² − 28 = 0 ;
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Divide the entire equation by "7" (both sides), to simplify;
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→ (7q² − 28) / 7 = 0 / 7 ;
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To get: → q² − 4 = 0 ;
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We can factor: " q² − 4 " ; into: "(q +2)*(q− 2)" ;
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and rewrite: q² − 4 = 0 ;
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as : (q +2)(q− 2) = 0 ;
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So, since anything multiplied by "0" equals "0" ;
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When "(q+2)" = 0 ; the equation holds true.
So, q + 2 = 0 ;
→ Subtract "2" from EACH side of the equation:
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q + 2 − 2 = 0 − 2 ;
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to get: q = -2 ;
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As such, when "(q−2)" = 0, the equation holds true.
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So, q − 2 = 0 ;
Add "2" to both sides of the equation:
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q − 2 + 2 = 0 + 2 ;
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to get: q = 2 .
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So we have 2 (TWO) values for "q" —which are "2" and "-2" .
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So the solution set is: q = {2, -2) ; [or, write as: q = {± 2} ] .
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