1. ## quadratic algebra

hi all,

the question posed is

Given that $\displaystyle x^2-14x+a = (x+b)^2$ for all values of $\displaystyle x$, find the value of $\displaystyle a$ and the value of $\displaystyle b$.

is this correct?

since

$\displaystyle (x+A)^2=x^2+2Ax+A^2$

$\displaystyle x^2+2Ax=(x+A)^2-A^2$

so

$\displaystyle x^2-14x = (x+b)^2 - b^2 = x^2+2bx$

and solving for b

$\displaystyle b=\frac{-14x}{2x}=-7$

so

$\displaystyle x^2-14x+a=(x-7)^2$

and expanding the RHS

$\displaystyle a=49$

2. Originally Posted by sammy28
hi all,

the question posed is

Given that $\displaystyle x^2-14x+a = (x+b)^2$ for all values of $\displaystyle x$, find the value of $\displaystyle a$ and the value of $\displaystyle b$.

is this correct?
...
solving

$\displaystyle b=-7$

$\displaystyle a=49$
That's the same as what I got.

3. Your solution is correct, however there is a much easier way of getting to it:

Simply look at the equation when $\displaystyle x=0,-b$ (you can do this since you are told that it holds for any $\displaystyle x$

Then, you get: $\displaystyle x=0$:

$\displaystyle 0^2 -14*0 + a = (0 + b)^2 \Rightarrow a = b^2$

$\displaystyle x=-b$:
$\displaystyle (-b)^2 +14b + a = (b-b)^2 = 0 \Rightarrow b^2 + 14b + b^2 = 0 \Rightarrow$
$\displaystyle \Rightarrow 2b^2 + 14b = 0 \Rightarrow 2b(b+7) = 0 \Rightarrow b = 0,-7$

This gives us two possible solutions, however $\displaystyle a=b=0$ is obviously wrong! so we are left with $\displaystyle \boxed{b = -7, a = 49}$

4. Since two quadratic expressions,

$\displaystyle x^2+2bx+b^2$

and
$\displaystyle x^2-14x+a$ are identical,
their coefficients will be in proportion.

i.e. $\displaystyle 1=-14/2b = (b^2)/a$

so $\displaystyle 2b=-14 and a=b^2$

So$\displaystyle b=-7 , a=49$

5. thanks for the replies. its good to see it approached from different angles.

defunkt i like your logic!

thanks a lot
sammy