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# Another P1 thing watch

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1. m and n are negative constants such that

(m - √3)(1 + n√3) = 5√3 - 7

Find the exact values of m and n.

I tried multiplying out, then realised that was going the wrong way and that they were identities, but I'm used to adding ones. So I put the first bracket equal to 1, and got a value for n, then realised it said "m and n are...constants", and screamed. So I gave multiplying out a go, then realised equating coefficients is again what you do when the letters are variables.

Help!
2. (Original post by mik1a)
m and n are negative constants such that

(m - √3)(1 + n√3) = 5√3 - 7

Find the exact values of m and n.

I tried multiplying out, then realised that was going the wrong way and that they were identities, but I'm used to adding ones. So I put the first bracket equal to 1, and got a value for n, then realised it said "m and n are...constants", and screamed. So I gave multiplying out a go, then realised equating coefficients is again what you do when the letters are variables.

Help!
You are going about this the right way.

I am probably wrong here but I think the answers are:

m = [(7-√109)/2]-7

n = (7-√109)/6

They just don't seem nice answers
3. Forget those answers. They're utter crap.
4. (Original post by mik1a)
m and n are negative constants such that

(m - √3)(1 + n√3) = 5√3 - 7

Find the exact values of m and n.

I tried multiplying out, then realised that was going the wrong way and that they were identities, but I'm used to adding ones. So I put the first bracket equal to 1, and got a value for n, then realised it said "m and n are...constants", and screamed. So I gave multiplying out a go, then realised equating coefficients is again what you do when the letters are variables.

Help!
expanding gives m+mnsqrt3-sqrt 3 -3n so clearly we know that
m-3n=7 and mn-1=5
so m=7+3n
n(7+3n)=5
=>3n^2+7n-5=0
n=-7+-sqrt(49+4*5*3)/6=(-7+-sqrt109)/6??? and then u can get m?
i think i have gone wron sumwheer
5. I tried it, but then got (m- sqrt3)(1-1) = 5 sqrt 3 -7

and it doesn't work from thereon.
6. (Original post by mik1a)
m and n are negative constants such that

(m - √3)(1 + n√3) = 5√3 - 7

Find the exact values of m and n.

I tried multiplying out, then realised that was going the wrong way and that they were identities, but I'm used to adding ones. So I put the first bracket equal to 1, and got a value for n, then realised it said "m and n are...constants", and screamed. So I gave multiplying out a go, then realised equating coefficients is again what you do when the letters are variables.

Help!
Ok i think i've managed it.. reasonable sounding answers.

(m - √3)(1 + n√3) = 5√3 - 7

m + mn√3 - √3 - n√3(√3) = 5√3 - 7

(mn√3 - √3) + (m - 3n) = 5√3 - 7

terms involving √3 equal..

therefore eqn1) 5√3 = mn√3 - √3

and eqn 2) -7 = m - 3n

so using 2) m = 3n - 7

sub into 1) 6√3 = (3n - 7)n√3

so 6 = -7n + 3n^2

solve quadratic.. (3n + 2)(n - 3) = 0

n = 3 OR n = -2/3 ... we're told its negative so n = -2/3.

then using 2) m = 3n - 7
sub in vals of n..
m = -9 OR m = 2

so n = -2/3 , m = -9
7. (Original post by lgs98jonee)
expanding gives m+mnsqrt3-sqrt 3 -3n so clearly we know that
m-3n=7 and mn-1=5
so m=7+3n
n(7+3n)=5
=>3n^2+7n-5=0
n=-7+-sqrt(49+4*5*3)/6=(-7+-sqrt109)/6??? and then u can get m?
i think i have gone wron sumwheer
so m=7+(-7+-sqrt109)/6
=(35+-sqrt109)/6
8. (Original post by Sahir)
so n = -2/3 , m = -9
that's what i got
9. (Original post by mockel)
that's what i got
Sounds good to me.
10. (Original post by Sahir)
Ok i think i've managed it.. reasonable sounding answers.

(m - √3)(1 + n√3) = 5√3 - 7

m + mn√3 - √3 - n√3(√3) = 5√3 - 7

(mn√3 - √3) + (m - 3n) = 5√3 - 7

terms involving √3 equal..

therefore eqn1) 5√3 = mn√3 - √3

and eqn 2) -7 = m - 3n

so using 2) m = 3n - 7

sub into 1) 6√3 = (3n - 7)n√3

so 6 = -7n + 3n^2

solve quadratic.. (3n + 2)(n - 3) = 0

n = 3 OR n = -2/3 ... we're told its negative so n = -2/3.

then using 2) m = 3n - 7
sub in vals of n..
m = -9 OR m = 2

so n = -2/3 , m = -9
yes!
11. (Original post by Sahir)
Ok i think i've managed it.. reasonable sounding answers.

(m - √3)(1 + n√3) = 5√3 - 7

m + mn√3 - √3 - n√3(√3) = 5√3 - 7

(mn√3 - √3) + (m - 3n) = 5√3 - 7

terms involving √3 equal..

therefore eqn1) 5√3 = mn√3 - √3

and eqn 2) -7 = m - 3n

so using 2) m = 3n - 7

sub into 1) 6√3 = (3n - 7)n√3

so 6 = -7n + 3n^2

solve quadratic.. (3n + 2)(n - 3) = 0

n = 3 OR n = -2/3 ... we're told its negative so n = -2/3.

then using 2) m = 3n - 7
sub in vals of n..
m = -9 OR m = 2

so n = -2/3 , m = -9
I've checked that and it is right. Well done.
12. (Original post by lgs98jonee)
yes!
cant believe i didnt get that right i put m=PLUS7PLUS 3n and not MINUS 7...i can believe how dumn i was :-(((
13. So you need to equate coefficients of √3? I suppose that's fair since it's an irrational number, no other number could be exactly that.
14. Find the exact square roots of 11+2√30 in surd form.

I don't know how you take the square root of (a+b), or at least I don't think I do. I know it's not the same as √a+√b at least.
15. So it is asking you to find the square root of 11 + 2 sqrt30 ?
16. (Original post by mik1a)
Find the exact square roots of 11+2√30 in surd form.

I don't know how you take the square root of (a+b), or at least I don't think I do. I know it's not the same as √a+√b at least.
how do u get the sqrt sign?
17. 11+2√30 =sqrt(121+120)
=sqrt(241)
and sqrt of that =241^0.25
18. microsoft word lets you use symbols, then I copy and paste. there's probably a alt+number combination for it as well.

2276: yeah I guess so from what the questions says.
19. I am afraid you're wrong lgs98jonee

241^0.25 = 3.94007293....

√(11+2√30) = 4.68555772....
20. (Original post by lgs98jonee)
11+2√30 =sqrt(121+120)
=sqrt(241)
and sqrt of that =241^0.25
√a + √b = √(a+b) is incorrect.

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