SOLUTION: Find the range of values of p for which 4x^2+12x+15=p(4x+5) has real roots.

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Question 1050486: Find the range of values of p for which 4x^2+12x+15=p(4x+5) has real roots.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Use the discriminant,
4x%5E2%2B12x%2B15=4px%2B5p
4x%5E2%2B12x-4px%2B15-5p=0
4x%5E2%2B%2812-4p%29x%2B%2815-5p%29=0
So then,
D=b%5E2-4ac
D=%2812-4p%29%5E2-4%2815-5p%29%284%29
For real roots,
D%3E=0
+%2812-4p%29%5E2-16%2815-5p%29%3E=0+
%28144-96p%2B16p%5E2%29-240%2B80p%3E=0
16p%5E2-16p-96%3E0
p%5E2-p-6%3E=0
%28p-3%29%28p%2B2%29%3E=0
Use the critical points, p=3 and p=-2,
Break up the number line into 3 regions,
Region 1 : p%3C-2
Region 2 : -2%3Cp%3C3
Region 3 : p%3E3
Choose a value in each region and check the inequality,
Region 1 : p=-3
%28-3-3%29%28-3%2B2%29%3E=0
-6%28-1%29%3E=0
6%3E=0
True
Region 2 : p=0
%280-3%29%280%2B2%29%3E=0
-6%3E=0
False
Region 3 : p=4
%284-3%29%284%2B2%29%3E=0
1%286%29%3E=0
6%3E=0
True
So then,
p%3C=-2Up%3E=3