Equation Help: Find the Values of a & b in (x+a)² + b = x² - 6x + 13

[zandra_z]

If (x + a)² + b = x² - 6x + 13
find the values of a and b

there are 3 unknowns and this confuses me... please help

 

[shmoe]

You have a polynomial on the left hand side and a polynomial on the right hand side. Two polynomials are equal if and only if their coefficients match.

eg.if 2x^2-3x+4=ax^2+bx+c then we must have a=2,b=-3,c=4

For your question, try expanding the (x+a)^2 part and equating coefficients.

 

[zandra_z]

x² + 2ax + a² + b = x² - 6x + 13
the x² cancel out
a² + 2ax + b = 13 - 6x

then I get stuck

 

[shmoe]

x² + 2ax + a² + b = x² - 6x + 13
the x² cancel out
a² + 2ax + b = 13 - 6x

then I get stuck

Collect like terms (according to power of "x"). Your equation becomes:

(2a)x+(a^2+b)=-6x+13

Switching the order of the left hand side was just to make things match up nicer. Remember what I said about the coefficients of equal polynomials.

2a=??
a^2+b=??

Can you fill in the ??

 

[arildno]

Write your last equation as:
(2a+6)x=13-a^2-b
Since this equation shall hold for ALL choices of x, in particular it must hold for x=0
which means you get the equations:
0=13-a^2-b
And:
2a+6=0

 

[zandra_z]

so:
a=-3
and:
0=16^2-b
b=2

 

[arildno]

How do you get that flawed b-value?

 

[zandra_z]

I see, I did:
16^(2-b)
I should do:
13-(3^2)-b
b=4