QUADRATIC
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➣ Problem:
Find the solution set of the quadratic equation $ \sf (t+5)^2 - 7(t+5)+6 = 0 $
➣ Answer:
The solution set is $ \sf \{ 1, -4 \} $.
➣ Solution:
» Transform first to standard form.
- $ \sf (t+5)^2 - 7(t+5)+6 = 0 $
- $ \sf (t^2 + 10t + 25) + (- 7t - 35) + 6 = 0 $
- $ \sf t^2 + 10t + 25 - 7t - 35 + 6 = 0 $
- $ \sf t^2 + (10t - 7t) + (25 - 35 + 6) = 0 $
- $ \sf t^2 + 3t + (-4) = 0 $
- $ \sf t^2 + 3t -4 = 0 $
» Factor.
- $ \sf t^2 + 3t -4 = 0 $
- $ \sf t^2 + (-t + 4t) -4 = 0 $
- $ \sf (t^2 - t) + (4t -4) = 0 $
- $ \sf t (t - 1) + 4 (t - 1) = 0 $
- $ \sf (t - 1)(t + 4) = 0 $
» Equate each of the two factors to zero and solve for x.
- $ \sf (t - 1)(t + 4) = 0 $
- $ \sf (t - 1) = 0 \\ \sf t = 1 $
- $ \sf (t + 4) = 0 \\ \sf t = -4 $
Thus, the solution set is $ \sf \{ 1, -4 \} $.
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