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# Maths question, help needed watch

1. Not sure if that question is right.

I would complete the square of LHS.

so: (x - 3)² - 9 +a

That would give b = -3

The RHS of your equation doesnt contain an 'a'. So a= 9
2. (Original post by Narin)
Expanding (x + b)^2 you get, x^2 + 2bx + b^2

Therefore:

2b = 6 and b^2 = a

From there it's an easy solve to get b=3 and a=9

i see what you mean, so you make 2bx equal to 6x from the other side of the equation, hence 2b =6, and b=3...

but that doesnt really prove that b=3, it just fits in with the equation so we're putting a 6x on both sides... we dont know that b=6 just cos of that..
i realise that what i just said makes no sense whatsoever by the way, but its hard explaining maths
3. Sorry.. i made a mistake.. 2b=-6 ---> b=-3, a still equals 9

i see what you mean, so you make 2bx equal to 6x from the other side of the equation, hence 2b =6, and b=3...

but that doesnt really prove that b=3, it just fits in with the equation so we're putting a 6x on both sides... we dont know that b=6 just cos of that..
i realise that what i just said makes no sense whatsoever by the way, but its hard explaining maths
I don't really see what your point is. The question is just find the values of a and b, and my values work in the equation:

(x + -3)^2 = x^2 -3x - 3x + 9 = x^2 -6x + 9

There's nothing else it could be.
4. Ok i see it now, cheers
5. (Original post by Narin)
Sorry.. i made a mistake.. 2b=-6 ---> b=-3, a still equals 9

I don't really see what your point is. The question is just find the values of a and b, and my values work in the equation:

(x + -3)^2 = x^2 -3x - 3x + 9 = x^2 -6x + 9

There's nothing else it could be.
yeh sorry i'm not making much sense, but what i'm saying is that how do we know there are no other answers it could be? cos we assume that b² = a , and that -6x = 2bx , when that's not necessarily true ... but i understand that's the answer they're looking for, because there's no other way of solving it given the information in the question.. but i dont think its true to say that just cos

-6x + a = 2bx + b²

that we have to make -6x = 2bx and do it like that, and thus make 2b =-6, thats not necessarily the only answer, though its the only one they can expect us to get..

cheers for your help
6. It is the only answer! It's kinda hard to explain, but trust me, there are no other solutions.
7. I would complete the square of LHS.

so: (x - 3)² - 9 +a

That would give b = -3

The RHS of your equation doesnt contain an 'a'. So a= 9
8. (Original post by Narin)
It is the only answer! It's kinda hard to explain, but trust me, there are no other solutions.
but because we dont know what X is, couldnt it be that 2bx is actually equal to - 8x, for example, and that b² and a are NOT equal.. its certainly possible.

eg the equation goes to this, as we've established:

-6x + a = 2bx + b²

but we can't prove there isnt a solution which goes, for example

-6x + a = -8x +b²

we are making the assumption that -6x equals 2bx and that a equals b², though i appreciate thats what the exam wants us to do. its certainly not true to say those are the only answers though, just the only ones we can find without trial and error.

anyway im gona forget about it now, and thanks for the help again
9. (Original post by hellohello)
but because we dont know what X is, couldnt it be that 2bx is actually equal to - 8x, for example, and that b² and a are NOT equal.. its certainly possible.

eg the equation goes to this, as we've established:

-6x + a = 2bx + b²

but we can't prove there isnt a solution which goes, for example

-6x + a = -8x +b²

we are making the assumption that -6x equals 2bx and that a equals b², though i appreciate thats what the exam wants us to do. its certainly not true to say those are the only answers though, just the only ones we can find without trial and error.

anyway im gona forget about it now, and thanks for the help again
Surely my way is,well, just easier.
10. (Original post by !Laxy!)
Surely my way is,well, just easier.
yeh, it is, sorry didnt read it properly. cheers
11. GAHHHH!! Stop over-complicating this!!

I highly doubt there is a way of making x^2 + bx + c = x^2 + dx + e, (for all values of x)which is what you are saying when to say that two equal quadratic equations, the values of the coefficients do not necessarily have to be equal.
12. x² - 6x + 9 = (x-3)²
13. I would complete the square of LHS.

so: (x - 3)² - 9 +a

That would give b = -3

The RHS of your equation doesnt contain an 'a'. So a= 9

SEE I DID GET 'a' AND ITS SIMPLE
14. (Original post by !Laxy!)
I would complete the square of LHS.

so: (x - 3)² - 9 +a

That would give b = -3

The RHS of your equation doesnt contain an 'a'. So a= 9

SEE I DID GET 'a' AND ITS SIMPLE
yeh sorry i misread it, and it is simple, cheers.
this question is well and trully exhausted, how do i close this thread
15. ok. i think this is the answer: -

x² - 6x + a = (x + b)² - similar to completing the square.

you have to find a such that it will give a repeated root, i.e.
(x + b) x (x + b)
so, when you write it out, you get:

x² + 2bx + b² - 2b has to give -6 so b is -3

b² is a (if you look at the left side of the equation)
-3² gives 9

so a = 9 and b = -3
16. (Original post by icecreamman)
ok. i think this is the answer: -

x² - 6x + a = (x + b)² - similar to completing the square.

you have to find a such that it will give a repeated root, i.e.
(x + b) x (x + b)
so, when you write it out, you get:

x² + 2bx + b² - 2b has to give -6 so b is -3

b² is a (if you look at the left side of the equation)
-3² gives 9

so a = 9 and b = -3
Grrrrrrrrr! Just complete the dam square! Its so much easier.
17. oh damn, you figured it out. my one moment of glory was taken away from me.
18. (Original post by Narin)
It is the only answer! It's kinda hard to explain, but trust me, there are no other solutions.

Alright sorry, I just have to point out one last thing.

x² - 6x + a = (x+b)²

find a and b, thats the original question

if i said to you X = 10, a = 185 and b=5

substitue those numbers into the equation above, and it works. you get 225=225.

hence proving that "a" could = 185, and b could = 5,
note that this IS a different solution
there are plenty of other solutions, and i've just proved than a does not have to equal 9, and b does not have to equal -3
but as i said, i appreciate those are the answers the examiners are looking for
19. It's only a solution if x=10, the correct answer works for any real value of x.
20. (Original post by hellohello)
Ok i see it now, cheers
wow, you see it? That's amazing, because when you first posted the question, I knew the answer, but after reading all the replies to your question, I feel thoroughly confused. Talk about wandering round and round in circles to get a simple answer...

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