This discussion is now closed.

Check out other Related discussions

Help with homework(rep is awarded)!!

Given that the point with coordinates (1 +√3 , 5√3 ) lies on the curve with the equation
y = 2² + px + q,
find the values of the rational constants p and q.

Reply 1

Jonny W

8

5sqrt(3)
= 2(1 + sqrt(3))^2 + p(1 + sqrt(3)) + q
= 2(4 + 2sqrt(3)) + p(1 + sqrt(3)) + q
= 8 + p + q + sqrt(3) (4 + p)

0 = 8 + p + q
5 = 4 + p

p = 1
q = -9

Reply 2

the bear

20

IZZY!

Given that the point with coordinates (1 +√3 , 5√3 ) lies on the curve with the equation
y = 2² + px + q,
find the values of the rational constants p and q.

hi izzy did you mean

y = 2x² + px + q ?

do i get a rep ?

ther bear

Reply 3

yazan_l

5

the bear

hi izzy did you mean

y = 2x² + px + q ?

do i get a rep ?

ther bear

i guess the rep is for the one who does the homework,,, not for the one one gives him the homework again!!! :rolleyes:

Reply 4

the bear

20

it was worth a try !!

Reply 5

Eddie K

2

Worked Solution:

y = 2x2 + px + q

When x=(1+√3) and y=(5√3):

5√3 = 2(1+√3)² + p(1+√3) + q
5√3 = 2(1+√3)(1+√3) + p(1+√3) + q
5√3 = 2[1+2√3+3] + p(1+√3) + q
5√3 = 2 + 4√3 + 6 + p + √3.p + q
5√3 = 8 + 4√3 + p + √3.p + q
5√3 = √3(4+p) + 8 + p + q

Equating constants:

5 = 4 + p
p = 1

and:

0 = 8 + p + q

Substitute p:

0 = 8 + 1 + q
0 = 9 + q
q = -9

p = 1, q = -9

Reply 6

IZZY!

OP

15

i repped the first person to reply. i ll rep again if some1 explains why do you get
5√3 = √3(4+p) + 8 + p + q
and then
5 = 4 + p

thanks.
the bear remind me later, like in 2 days and i ll rep u lol.

Reply 7

Eddie K

2

IZZY!

i repped the first person to reply. i ll rep again if some1 explains why do you get
5√3 = √3(4+p) + 8 + p + q
and then
5 = 4 + p

thanks.
the bear remind me later, like in 2 days and i ll rep u lol.

The left hand side must equal the right hand side, as with all equations.

By inspection, it is clear that there is a 'coefficient' for √3 of 5 on the left hand side. On the right hand side, there is also a 'coefficient' of √3 of (4+q). These must be equal, so you get:

5 = 4 + p

There is nothing else on the left hand side but still 8 + p + q on the right hand side, so therefore it follows that also:

0 = 8 + p + q

Reply 8

the bear

20

IZZY!

i repped the first person to reply. i ll rep again if some1 explains why do you get
5√3 = √3(4+p) + 8 + p + q
and then
5 = 4 + p

thanks.
the bear remind me later, like in 2 days and i ll rep u lol.

it's in my diary !!

Reply 9

steve2005

17

IZZY!

like in 2 days and i ll rep u lol.

I don't understand the end bit .

I presume 'rep' means reputation and ' u ' means you : that's easy but what about ' lol.'

Reply 10

Gaz031

15

steve2005

I don't understand the end bit .

I presume 'rep' means reputation and ' u ' means you : that's easy but what about ' lol.'

I'm fairly sure that 'lol' is supposed to be an abbreviation for 'laugh out loud' though the vast majority of people don't actually laugh at all when they say 'lol'.

Reply 11

steve2005

17

Gaz031

I'm fairly sure that 'lol' is supposed to be an abbreviation for 'laugh out loud' though the vast majority of people don't actually laugh at all when they say 'lol'.

Thanks

Reply 12

IZZY!

OP

15

please help again.solve this.
a Show that
(x² + 2x − 3)(x² − 3x − 4) ≡ x⁴ − x³ − 13x² + x + 12.
b Hence solve the equation
x⁴ − x³ − 13x² + x + 12 = 0.

i ve done the a part.i cant do the b part.help is greatly appreciated and will be awarded with rep, if you remind me later.

Reply 13

Gaz031

15

IZZY!

please help again.solve this.
a Show that
(x² + 2x − 3)(x² − 3x − 4) ≡ x⁴ − x³ − 13x² + x + 12.
b Hence solve the equation
x⁴ − x³ − 13x² + x + 12 = 0.

i ve done the a part.i cant do the b part.help is greatly appreciated and will be awarded with rep, if you remind me later.

For part (b) use the result obtained in part (a) and the fact that if f(x)g(x)=0 then f(x)=0 or g(x)=0 (or both).

Reply 14

Chewwy

17

IZZY!

please help again.solve this.
a Show that
(x² + 2x − 3)(x² − 3x − 4) ≡ x⁴ − x³ − 13x² + x + 12.
b Hence solve the equation
x⁴ − x³ − 13x² + x + 12 = 0.

i ve done the a part.i cant do the b part.help is greatly appreciated and will be awarded with rep, if you remind me later.

having done a by expanding it, you know x^4... is divisible by those 2 factors. hmm. but can those factors also be broken down into factors? this will give you the fully factorised expression of x^4... and you should know how to find the roots from that.

Reply 15

IZZY!

OP

15

thanks chewyy, u r on my replist.

Help with homework(rep is awarded)!!

Related discussions

Articles for you