# Maths algebra help please !!!!!!Watch

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#1
Hi. I was wondering if anybody could help me to understand this question, if you could that would help a lot many thanks!!

Work out the value of m, k and n:

2x^2+8x+k=m(x-n)^2+5n

M=2 n=-2 k=-2

And I am completely confused!
0
2 years ago
#2
(Original post by Snowystar)
Hi. I was wondering if anybody could help me to understand this question, if you could that would help a lot many thanks!!

Work out the value of m, k and n:

2x^2+8x+k=m(x-n)^2+5n

M=2 n=-2 k=-2

And I am completely confused!
Better off posting this directly in the maths forum where you will get a quick response.
0
2 years ago
#3
(Original post by Snowystar)
Hi. I was wondering if anybody could help me to understand this question, if you could that would help a lot many thanks!!

Work out the value of m, k and n:

2x^2+8x+k=m(x-n)^2+5n

M=2 n=-2 k=-2

And I am completely confused!
Hiya

For this question it's best to look at the differences on both sides, on the left you have a standard quadratic, and on the right you have a quadratic (with the coefficients missing) written as you would when you are completing the square.

Does this make it any clearer?
0
2 years ago
#4
You have on quadratic in x equal to another. For them to be equal for all values of x, the coefficients of each of the terms (x^2, x and constant) must be equal. Expand the RHS and you will be able to calculate the values of the constants k, m and n.

To get started, the coefficient of x^2 on the LHS is 2. If you expand the RHS, its x^2 coefficient is m, so m=2. Do the same the the x and constant terms.
1
#5
(Original post by KaylaB)
Hiya

For this question it's best to look at the differences on both sides, on the left you have a standard quadratic, and on the right you have a quadratic (with the coefficients missing) written as you would when you are completing the square.

Does this make it any clearer?
Yes I see now! Thank you!
0
2 years ago
#6
(Original post by Snowystar)
Yes I see now! Thank you!
No problem, if you have any further issues with it just give me a shout
0
2 years ago
#7
carefully remove the brackets on the RHS & compare the various powers of x on both sides.
0
#8
(Original post by Snowystar)
Yes I see now! Thank you!
Wait I still don't understand how to get k though
0
2 years ago
#9
(Original post by Snowystar)
Wait I still don't understand how to get k though
So if you now know the values of m and n (m = 2 and n = -2)
You can find k by recognising that -8 + k = 5n
And since you know the value of n you can just substitute that value in and find k
1
#10
(Original post by KaylaB)
So if you now know the values of m and n (m = 2 and n = -2)
You can find k by recognising that -8 + k = 5n
And since you know the value of n you can just substitute that value in and find k
Ok got it is the following question right?

X^2-6x+k=m(x-n)^2+2n
Does m=1 n=3 and k=-6
?
0
2 years ago
#11
(Original post by Snowystar)
Ok got it is the following question right?

X^2-6x+k=m(x-n)^2+2n
Does m=1 n=3 and k=-6
?
Not quite, do you mind posting your working so I can help guide you through it?
0
#12
(Original post by KaylaB)
Not quite, do you mind posting your working so I can help guide you through it?
No i don't mind. I really appreciate the help, thank you 😊

X^2-6x+k=m(x-n)^2+2n
So straight away m=1

Then i deal with the left of the equation to get (x-3)^2 +k or (x-3)^2 +9
So n must be 3
So now the right hand side
(X-3)^2 +9+(2x3)
To get back to 9 do -6 so k is -6

There's my working i actually have no idea why i done the things that I done but ...
0
2 years ago
#13
(Original post by Snowystar)
So now the right hand side
(X-3)^2 +9+(2x3)
To get back to 9 do -6 so k is -6

There's my working i actually have no idea why i done the things that I done but ...
Where did the +9 come from?

I find it easier to multiply-out the RHS, rather than completing the square on the LHS.
0
2 years ago
#14
(Original post by Snowystar)
No i don't mind. I really appreciate the help, thank you 😊

X^2-6x+k=m(x-n)^2+2n
So straight away m=1

Then i deal with the left of the equation to get (x-3)^2 +k or (x-3)^2 +9
So n must be 3
So now the right hand side
(X-3)^2 +9+(2x3)
To get back to 9 do -6 so k is -6

There's my working i actually have no idea why i done the things that I done but ...
You're doing the correct method except for it would be -9 instead of +9 (as if you expanded (x-3)^2 you have an extra 9) , and when you were finding k for this question x^2-6x+k=m(x-n)^2+2n you should be able to see that the bits in bold below are equal to each other, and since you know the value of n you can just do -9+k = 2n
(x-3)^2 -9+k = m(x-n)^2+2n

But you're applying the correct working to a question which is different to the one you originally posted! 2x^2+8x+k = m(x-n)^2+5n
If you apply the same method and understanding to the question above you should get the right answers
0
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