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quadratic inequality

Usually sound at these, dunno if it's me or the question, idk what to do next:






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Reply 1
Avatar for Hasufel
Hasufel
16
if it has 2 real roots, then great, but the question is indicating that the roots might also be equal , so the "greater than or equal to" in the discriminant applies.

i.e. (k−1)2−16≥0,=>(k−1+4)(k−1−4)≥0(k-1)^2-16 \geq 0 , => (k-1+4)(k-1-4) \geq 0

(difference of 2 squares)

(EDITED - mistake inside brackets!!!)
(edited 9 years ago)
Reply 2
Avatar for HeadyKiss
HeadyKiss
OP
2
Original post by Hasufel
if it has 2 real roots, then great, but the question is indicating that the roots might also be equal , so the "greater than or equal to" in the discriminant applies.

i.e. (k−1)2−16≥0,=>(k−1+16)(k−1−16)≥0(k-1)^2-16 \geq 0 , => (k-1+16)(k-1-16) \geq 0

(difference of 2 squares)

(EDITED)


...I've just spotted I misread the question as "only one real root" :colondollar: Thanks

Spoiler

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Reply 3
Avatar for Hasufel
Hasufel
16
you spottedi t even with my (rushed - i`m doing 1000 other thinks, apologies) stupid-ass mistake!!
(edited 9 years ago)

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