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FP2 inequality question

Can't see my mistake.

Find the set of values of y, for which there are no points on the curve y=4x−3a2(x2+a2) y=\frac{4x-3a}{2(x^2+a^2)} where a is a positive constant.

So I rearranged into a quadratic and used the discriminant to get to:

4≥2ay[2ay+3] 4 \geq 2ay[2ay+3] . So then

2a≥y \frac {2}{a} \geq y so y < −2a -\frac{2}{a} .

Using the same logic I did this to the other factor and end up with 12a≥y \frac{1}{2a}\geq y but the markscheme gets y> 1/2a
Reply 1
Avatar for TeeEm
TeeEm
19
my discriminant gets 2ay^2+3a^2y-2>0

which I get y<-2/a or y>1/(2a)
Reply 2
Avatar for TeeEm
TeeEm
19
CORRECTION my discriminant is 2(ay)^2+3a^2y-2>0
Reply 3
Avatar for davros
davros
16
Original post by maths learner
Can't see my mistake.

Find the set of values of y, for which there are no points on the curve y=4x−3a2(x2+a2) y=\frac{4x-3a}{2(x^2+a^2)} where a is a positive constant.

So I rearranged into a quadratic and used the discriminant to get to:

4≥2ay[2ay+3] 4 \geq 2ay[2ay+3] . So then

2a≥y \frac {2}{a} \geq y so y < −2a -\frac{2}{a} .

Using the same logic I did this to the other factor and end up with 12a≥y \frac{1}{2a}\geq y but the markscheme gets y> 1/2a


You're doing something very funny with inequalities there - if 2/a >= y then y <= 2/a. You seem to have got the right answer with the wrong method!

Original post by TeeEm
CORRECTION my discriminant is 2(ay)^2+3a^2y-2>0


My discriminant is 2a^2y^2 + 3ay - 2 > 0 :smile:

We can divide by 2a^2 since a^2 is a positive constant to get

y^2 + (3y/2a) - (1/a^2) > 0

and then completing the square and rearranging leads to 2 inequalities y > 1/(2a) or y < (-2/a) which hopefully are what the mark scheme gives!
(edited 9 years ago)
Original post by davros
You're doing something very funny with inequalities there - if 2/a >= y then y <= 2/a. You seem to have got the right answer with the wrong method!



My discriminant is 2a^2y^2 + 3ay - 2 > 0 :smile:

We can divide by 2a^2 since a^2 is a positive constant to get

y^2 + (3y/2a) - (1/a^2) > 0

and then completing the square and rearranging leads to 2 inequalities y > 1/(2a) or y < (-2/a) which hopefully are what the mark scheme gives!


Thanks for the reply. Yeah I have no idea what I was doing there! WIll teach me not to do questions when half asleep. Here is my working in full as I'm still making a mistake somewhere:

y=4x−3a2(x2+a2) y= \frac{4x-3a}{2(x^2+a^2)}
2y(x2+a2)=4x−3a 2y(x^2+a^2)=4x-3a
2yx2+2ya2−4x+3a=0 2yx^2+2ya^2-4x+3a=0
2yx2−4x+(2ya2+3a)=0 2yx^2-4x+(2ya^2+3a)=0

So now calculating my discriminant I get:

16−4[2y(2ya2+3a)]>0 16-4[2y(2ya^2+3a)]>0
4−4y2a2−6ay>0 4-4y^2a^2-6ay>0
2−2y2a2−3ay>0 2-2y^2a^2-3ay>0
2−ay(2ay+3)>0 2-ay(2ay+3)>0
2>ay(2ay+3) 2>ay(2ay+3)
2>ay=>y<2a 2>ay => y< \frac{2}{a}
2>2ay+3=>y<−12a 2>2ay+3 => y<-\frac{1}{2a}
Reply 5
Avatar for james22
james22
16
Original post by maths learner
Thanks for the reply. Yeah I have no idea what I was doing there! WIll teach me not to do questions when half asleep. Here is my working in full as I'm still making a mistake somewhere:

y=4x−3a2(x2+a2) y= \frac{4x-3a}{2(x^2+a^2)}
2y(x2+a2)=4x−3a 2y(x^2+a^2)=4x-3a
2yx2+2ya2−4x+3a=0 2yx^2+2ya^2-4x+3a=0
2yx2−4x+(2ya2+3a)=0 2yx^2-4x+(2ya^2+3a)=0

So now calculating my discriminant I get:

16−4[2y(2ya2+3a)]>0 16-4[2y(2ya^2+3a)]>0
4−4y2a2−6ay>0 4-4y^2a^2-6ay>0
2−2y2a2−3ay>0 2-2y^2a^2-3ay>0
2−ay(2ay+3)>0 2-ay(2ay+3)>0
2>ay(2ay+3) 2>ay(2ay+3)
2>ay=>y<2a 2>ay => y< \frac{2}{a}
2>2ay+3=>y<−12a 2>2ay+3 => y<-\frac{1}{2a}


The error is in your last two lines. You need to completely factorise the quadratic in y first.
Original post by james22
The error is in your last two lines. You need to completely factorise the quadratic in y first.


Thanks. I'm still making an error somewhere -.-.

0>y2+3y2a−1a2 0> y^2 + \frac{3y}{2a} -\frac{1}{a^2}
0>([y+34a)2]−916a2−1a2 0> ([y+\frac{3}{4a})^2]-\frac{9}{16a^2}-\frac{1}{a^2}
2516a2>(y+34a)2 \frac{25}{16a^2}>(y+\frac{3}{4a})^2
54a>±(y+34a) \frac{5}{4a}> \pm (y+\frac{3}{4a})
2>4ay=>12a>y=>y<12a... 2> 4ay => \frac{1}{2a}>y => y< \frac{1}{2a}...
Reply 7
Avatar for james22
james22
16
Original post by maths learner
Thanks. I'm still making an error somewhere -.-.

0>y2+3y2a−1a2 0> y^2 + \frac{3y}{2a} -\frac{1}{a^2}
0>([y+34a)2]−916a2−1a2 0> ([y+\frac{3}{4a})^2]-\frac{9}{16a^2}-\frac{1}{a^2}
2516a2>(y+34a)2 \frac{25}{16a^2}>(y+\frac{3}{4a})^2
54a>±(y+34a) \frac{5}{4a}> \pm (y+\frac{3}{4a})
2>4ay=>12a>y=>y<12a... 2> 4ay => \frac{1}{2a}>y => y< \frac{1}{2a}...


Again, your last 2 lines are wrong. With inequalities you cannot just square root both sides. Have you been shown how to solve quadratic inequalities before?

You need to factorise (not complete the square) first.
Original post by james22
Again, your last 2 lines are wrong. With inequalities you cannot just square root both sides. Have you been shown how to solve quadratic inequalities before?

You need to factorise (not complete the square) first.


Was never really taught inequalities properly. I'll look over it.

Factorise what exactly?
(edited 9 years ago)
Reply 9
Avatar for james22
james22
16
Original post by maths learner
Was never really taught inequalities properly. I'll look over it.

Factorise what exactly?


You need to factorise the quadratic in y, like you normally would do to solve it.
Original post by james22
You need to factorise the quadratic in y, like you normally would do to solve it.


Okay. I couldn't factorise it because of all the fractions so I used the quadratic formula.

y2+32ay−1a2<0 y^2 + \frac{3}{2a}y - \frac{1}{a^2} <0

Doing quadratic formula on this gave me:

−34a±54a-\frac{3}{4a} \pm \frac{5}{4a}
Original post by maths learner
Okay. I couldn't factorise it because of all the fractions so I used the quadratic formula.

y2+32ay−1a2<0 y^2 + \frac{3}{2a}y - \frac{1}{a^2} <0

Doing quadratic formula on this gave me:

−34a±54a-\frac{3}{4a} \pm \frac{5}{4a}


Since you know the solutions, you should now be able to factorise it because the factors will just be the solutions.
Original post by james22
Since you know the solutions, you should now be able to factorise it because the factors will just be the solutions.


Got it! Thanks so much. Reviewd quadratic inequalities as well :smile:.

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