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The equation |x^2px-51|=7 has real solutions
A. only when p^2 >/= 48
B. only when p^2< 48
C. for any p-value
D. for no values of p
A. only when p^2 >/= 48
B. only when p^2< 48
C. for any p-value
D. for no values of p
Answers
Answered by
oobleck
x^2+px-51 = ±7
so
x^2 + px - 58 = 0
p^2 + 232 >= 0
x^2 + px - 44 = 0
p^2 - 176 >= 0
None of that involves 48, so I suspect a typo.
You just need to recall that the discriminant b^2-4ac cannot be negative if the roots are real.
so
x^2 + px - 58 = 0
p^2 + 232 >= 0
x^2 + px - 44 = 0
p^2 - 176 >= 0
None of that involves 48, so I suspect a typo.
You just need to recall that the discriminant b^2-4ac cannot be negative if the roots are real.
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