# Whats the best way to approach a question such as:Watch

This discussion is closed.
#1
this one:

x^2+10x+36 == (x+a)^2+b

a, find the value of A and B

They was I approached this question, was by subsituting coefficient from the LHS to the RHS, like this

x^2+10x+36 == (x+a)(x+a)+b
x^2+10x+36 == x^2+ax+ax+a^2+b
x^2+10x+36 == x^2+2ax+a^2+b

therefore x^2 = 1

a = 2(x) ------- b = -a^2
a = 2(10) ------- b = -20^2
a = 20 -------- b = -400

Is this correct?

Thanks.
0
14 years ago
#2
Personally, I'm not sure about what you have done.

I would go about it like this;

xÂ²+10x+36=(x+a)Â²+b
=(x+5)(x+5)
=xÂ²+10x+25

[36-25=11]

So: (x+5)Â²+11=(x+a)Â²+b

a=5
b=11
0
14 years ago
#3
Do you know how to complete the square??

x^2+10x+36 == (x+a)^2+b

x^2 + 10x + 36 = (x+5)^2 - 25 + 36

= (x+5)^2 + 11

a=5 b =11
0
14 years ago
#4
firstly, x^2 is not 1. you are equating the coefficients only. equating those of x^2 tells you only that 1=1.

I'm not sure what you've done beyond that. you have to look at the corfficients of x. so you can say that:

10 = 2*a
a = 5

also, equating constants,

36 = a^2 + b

from this you can solve:

36 = 5^2 + b
b = 11
0
#5
(Original post by Slice'N'Dice)
Personally, I'm not sure about what you have done.

I would go about it like this;

xÂ²+10x+36=(x+a)Â²+b
=(x+5)(x+5)
=xÂ²+10x+25

[36-25=11]

So: (x+5)Â²+11=(x+a)Â²+b

a=5
b=11

I can see what you have done, however I am a bit confused on why you have factorised the equation x^2+10x+36 in the form (x+5)(x+5). Yes 5+5 goes into 10, but isnt 5x5 = 25 and not 36?

as for the rest of the equation I guess I will reamin confused until I understand what you did on your first step.

Thank you
0
14 years ago
#6
0
#7
(Original post by mik1a)
firstly, x^2 is not 1. you are equating the coefficients only. equating those of x^2 tells you only that 1=1.

I'm not sure what you've done beyond that. you have to look at the corfficients of x. so you can say that:

10 = 2*a
a = 5

also, equating constants,

36 = a^2 + b

from this you can solve:

36 = 5^2 + b
b = 11
Yup what you just did there is what i was tryign to do, I got in a bit of a muddle by the looks of it because for a for example 2ax, as they were all together I multiplied 2 by x to get 20, and so that how I came to the conclusion a was 20 and not 5.
0
14 years ago
#8
(Original post by ThugzMansion7)
Yep - the easiest way.
0
#9
(Original post by ThugzMansion7)
Completing the square is one area , I have never truely understood. Can someone here be kind enough to step by step tell me how to complete the square for that question. Sorry and thanks.
0
14 years ago
#10
(Original post by DOJO)

I can see what you have done, however I am a bit confused on why you have factorised the equation x^2+10x+36 in the form (x+5)(x+5). Yes 5+5 goes into 10, but isnt 5x5 = 25 and not 36?

as for the rest of the equation I guess I will reamin confused until I understand what you did on your first step.

Thank you

No problem. 5x5=25. Since you need 36, the difference between 36 and 25 is 11. [36-25=11]

(x+5)Â²=xÂ²+10x+25 Therefore you need to add 11 to get 36.

(x+5)Â²+11

a=5
b=11

Hope that is clear.

S'N'D.
0
14 years ago
#11
(Original post by DOJO)
I can see what you have done, however I am a bit confused on why you have factorised the equation x^2+10x+36 in the form (x+5)(x+5). Yes 5+5 goes into 10, but isnt 5x5 = 25 and not 36?

yes, you're right. the expression you are given is NOT a perfect square. whereas 9 and 16 can be written 3^2 and 4^2, 12 would have to be written as 3^2 + 3. some expressions ARE perfect squares. for example, x^2 + 6x + 9, which is the same as (x+3)^2.

however, some are not. when you factorise them, and multiply out, you have a difference - the factorised square and the original expression do not add up becuase the original expression was not a perfect square. this is why you have a "+b" on the outside of the square - this compensates for the difference.

so you can say:

x^2 + 10x + 36 DOES NOT equal : (x + 5)^2,
becuase (x+5)^2 = x^2 + 10x + 25.
the original expression was 11 more than the square. so because of this, you have to get the square, and add 11 to the outside of it to make its value upto the original one.

now you CAN say that:
x^2 + 10x + 36 = (x + 5)^2 + 11

because of that 11.
0
#12
(Original post by Slice'N'Dice)

No problem. 5x5=25. Since you need 36, the difference between 36 and 25 is 11. [36-25=11]

(x+5)Â²=xÂ²+10x+25 Therefore you need to add 11 to get 36.

(x+5)Â²+11

a=5
b=11

Hope that is clear.

S'N'D.

Oh I see now, so as a is in the brackets, it is '5' ....and the reason why you used 5 was because 5 goes into 10? I mean lets say if the equation was,

x^2+20+51 == (x+a)^2+b , then a would be 10 as (x+10)(x+10) which is
x^2+10x+10x+100 = 0 and so b would be 100-51 = 49?
0
14 years ago
#13
(Original post by DOJO)
Oh I see now, so as a is in the brackets, it is '5' ....and the reason why you used 5 was because 5 goes into 10? I mean lets say if the equation was,

x^2+20+51 == (x+a)^2+b , then a would be 10 as (x+10)(x+10) which is
x^2+10x+10x+100 = 0 and so b would be 100-51 = 49?
Yes that is correct. However b=-49 as you need 51. A positive 49 will give you 149 which is incorrect. Hope that is clear.

mik1a provided a detailed explanation about it.
0
#14
(Original post by Slice'N'Dice)
Yes that is correct. However b=-49 as you need 51. A positive 49 will give you 149 which is incorrect. Hope that is clear.

mik1a provided a detailed explanation about it.
TSR is great!, yeah I read that helped so much!

Why couldn't all text books explain completing the square like you guys have!
0
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