A maths - Surds, indices, roots

anpanman SG

Find the range of values of p for which the line y - x =2 meets the curve y^2 + (x+p)^2 = 2

My workings:

y = 2 + x (1)

y^2 +(x+p)^2 = 2 (2)

Sub (1) into (2)

2x^2 + (4 + 2p)x + 2 + p^2 = 0

Discriminant is more than or equal to 0

16 +16p + 4p^2 - 16 - 8p^2 more than or equal to 0

16p - 4p^2 > or equal to 0

.:p< or equal to 0 OR p> or equal to 4

My friend says the answer should be 0< or equal to P < or equal to 4. But I 'm pretty sure I did not make any mistakes in my working. Would someone please clarify this for me?

Thanks and Happy Chinese New Year!

jiaxing2 SG

"16p - 4p^2 > or equal to 0 "

if that is the case your friend is right.

since coefficient of p^2 is negative, graph is 'n' shape, not 'u' shape...thus p lies between and not beyond the intercept values.

jiaxing

anpanman SG

"16p - 4p^2 > or equal to 0 "

if that is the case your friend is right.

since coefficient of p^2 is negative, graph is 'n' shape, not 'u' shape...thus p lies between and not beyond the intercept values.

jiaxing

But if I were to factorise it I 'd get 4p(4-p) > or equal to 0

To determine whether p is beyond intercept point, we are required to sketch a small graph and since it is more than or equal to 0, we have to shade the top part of the graph and not between the values of 0 and 4.

Then shouldn't it be p< or equal to 0 or more than or equal to 4?

I might be wrong, but maybe you won't mind explaining it again?

thanks.

Lee012lee SG

The range of p values that correspond with the top part of the graph that is above the x-axis is 0 < p < 4.

Check your sketch graph, the graph is an " inverted U" graph, not a "U" graph.

SBS261P SG

your “top part of the graph” is between the values of zero and 4 =)

eagle SG

I can retire :D

Good good :D

Ajimal SG

Discriminant is more than or equal to 0

16 +16p + 4p² - 16 - 8p² ≥ 0

16p - 4p² ≥ 0

4p(4-p) ≥ 0

4p ≥ 0 and 4-p ≥ 0

p ≥ 0 -p ≥ -4

p ≥ 0 p ≤ 4 (Inequalities Rule: [ -x ≥ -y = x ≤ y ] Thus, -p ≥ -4 = p ≤ 4)

thus, 0 ≤ p ≤ 4 #

Looks like ur fren is rite :D. Hope this helped....

<EDIT>

Mikethm ask me to replace "or" with "and" :P... Thanks lol

Mikethm SG

Discriminant is more than or equal to 0

16 +16p + 4p² - 16 - 8p² ≥ 0

16p - 4p² ≥ 0

4p(4-p) ≥ 0

4p ≥ 0 or 4-p ≥ 0

p ≥ 0 -p ≥ -4

p ≥ 0 p ≤ 4 (Inequalities Rule: [ -x ≥ -y = x ≤ y ] Thus, -p ≥ -4 = p ≤ 4)

thus, 0 ≤ p ≤ 4 #

Looks like ur fren is rite :D. Hope this helped....

I believe your presentation and logic are wrong.

At step 4, you made the statement that

4p ≥ 0 or 4-p ≥ 0

Logically it imply that (at step 6)

p ≥ 0 or p ≤ 4

If p can be greater than zero OR less than 4. It imply that ONLY 1 condition need be satisfied. Meaning your next statement SHOULD be "all real values of p".

In addition, you neglected that if

both 4p≤0 and 4-p ≤0

Then 4p(4-p) ≥ 0 since -ve multiplied with -ve is +ve.

Thus to avoid such tedious thinking... please stick to critical region testing method or graphical method. If you should be planning to do your A levels, the graphical method is better as it train you to read solutions from a graph which is an essential skill at A level.

Ajimal SG

huh? so do u mean i should replace the "or" with "and" ? U could juz ask to replace "or" with "and" ..... Make life simple my fren :D ... Thanks for detailed explanations anyways...

Mikethm SG

huh? so do u mean i should replace the "or" with "and" ? U could juz ask to replace "or" with "and" ..... Make life simple my fren :D ... Thanks for detailed explanations anyways...

No, I just explaining that your working is wrong even if your answer is correct. To make life simple... don't write down the inequalities separately( aka step 4) as your reasoning is flawed. Draw a curve and derive the solution from the curve.

You should write

line 1/2/3

Followed by graph

Followed by solution(s) immediately.

Line (4) is the most common mistake students make in this topic.

SBS261P SG

haha, yep.

4p(4-p) ≥ 0

people usually think that both are positive (4p≥0, 4-p≥0) (CASE 1) but also can be both are negative (4p ≤ 0, 4-p ≤ 0)(CASE 2).

what Mikethm meant was, you only considered case 1, not case 2.

In short, your actual working should have been:

4p (4-p) ≥ 0

=> EITHER:

CASE 1: 4p ≥ 0 AND 4-p ≥ 0; OR

CASE 2: 4p ≤ 0 AND 4-p ≤ 0

then you solve for case one to get 0 ≤ p ≤ 4.

then you solve for case 2 to get p ≤ 0 AND p ≥ 4 (NOT APPLICABLE)

ONLY THEN will you get to the final answer 0 ≤ p ≤ 4.

a few marks lost…

SBS261P SG

so, in short, DON’T GO AND CONFUSE YOURSELF IF YOU DON’T WANT THE TROUBLE, use the tried and tested graph method