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c1 Algebra Relationships

Question: If x is real and p=3(x2+1)2x−1 p = \frac{3(x^2+1)}{2x-1} , prove that [br]p2−3(p+3)≥0 [br]p^2-3(p+3)\geq 0

My attempt:
p=3(x2+1)2x−1p = \frac{3(x^2+1)}{2x-1}

(2x−1)p=3(x2+1)(2x-1)p = 3(x^2+1)

2px−p=3x2+32px - p = 3x^2 + 3

3x2−2px+3+p=03x^2 - 2px + 3 + p = 0

∴b2−4ac=(−2p)2−4(3)(3+p)=4p2−12p−36 \therefore b^2 -4ac = (-2p)^2 - 4(3)(3+p) = 4p^2 -12p - 36 =4(p2−3p−9) =4(p^2-3p-9)

After this point i need help please
the problem is that if you let f(x)=4(p2−3p−9) f(x) = 4(p^2-3p-9)
if you complete the square of f(x) you find that the minimum value is -(45/4) which means that f(x) is not greater or equal to 0 for some values of x.
Due to this, how do i prove that (p2−3p−9)≥0 (p^2-3p-9)\geq 0

thanks in advance for helping me
(edited 7 years ago)
Reply 1
Avatar for Zacken
Zacken
22
Original post by bigmansouf
Question: If x is real and p=3(x2+1)2x−1 p = \frac{3(x^2+1)}{2x-1} , prove that [br]p2−3(p+3)≥0 [br]p^2-3(p+3)\geq 0

My attempt:
p=3(x2+1)2x−1p = \frac{3(x^2+1)}{2x-1}

(2x−1)p=3(x2+1)(2x-1)p = 3(x^2+1)

2px−p=3x2+32px - p = 3x^2 + 3

3x2−2px+3+p=03x^2 - 2px + 3 + p = 0

∴b2−4ac=(−2p)2−4(3)(3+p)=4p2−12p−36 \therefore b^2 -4ac = (-2p)^2 - 4(3)(3+p) = 4p^2 -12p - 36 =4(p2−3p−9) =4(p^2-3p-9)

After this point i need help please
the problem is that if you let f(x)=4(p2−3p−9) f(x) = 4(p^2-3p-9)
if you complete the square of f(x) you find that the minimum value is -(45/4) which means that f(x) is not greater or equal to 0 for some values of x.
Due to this, how do i prove that (p2−3p−9)≥0 (p^2-3p-9)\geq 0

thanks in advance for helping me


You know that xx is real, so the equation you have must have a discriminant greater than or equal to 0. So 4(p2−3p−9)≥04(p^2 - 3p - 9) \geq 0 and you're done, since your quadratic gives you solutions of the form x = something, but you're given that x is real, so your solutions to that quadratic is real, so the discriminant is >=0.
Reply 2
Avatar for bigmansouf
bigmansouf
OP
20
Original post by Zacken
You know that xx is real, so the equation you have must have a discriminant greater than or equal to 0. So 4(p2−3p−9)≥04(p^2 - 3p - 9) \geq 0 and you're done, since your quadratic gives you solutions of the form x = something, but you're given that x is real, so your solutions to that quadratic is real, so the discriminant is >=0.


im sorry can i ask another question please i know you are busy with your uni work
Reply 3
Avatar for Zacken
Zacken
22
Original post by bigmansouf
im sorry can i ask another question please i know you are busy with your uni work


about clarifying something I said? sure
Reply 4
Avatar for bigmansouf
bigmansouf
OP
20
Original post by Zacken
about clarifying something I said? sure


im sorry about i have being thinking about it all day and to be honest I didnt want to trouble you too much but it is irrating me so i have to ask. I know for you it may be trivial but i need it to be sortedin my mind
nomarlly in questions like this when x is real i assume that it means that the quadratic equation has two roots or an equal (repeated root). The problem i have in my head is that even though (p2−3p−9)≥0 (p^2-3p-9)\geq 0 the problem is that if p = 0 the quadratic equation 3x2+1 3x^2 + 1 which has no real roots. As a result, only some values of p will lead to the quadratic equation having real roots such as when p =4 creating 3x2−8x+4 3x^2 -8x +4 . The reason for thiniking this why is because in some A level inequality questions you are ask to find the range when the quadratic has real roots. Hence i was confused what to do when the question didnt ask for it since only some values of p will result in quadratic equations having real roots
Does what i say matter for this kind of question or am i thinking too much?

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