quadratic algebra

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August 28th 2009, 12:09 AM
sammy28

quadratic algebra

hi all,

the question posed is

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

is this correct?

since

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

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

so

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

and solving for b

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

so

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

and expanding the RHS

$a=49$
August 28th 2009, 01:26 AM
aidan

Quote:

Originally Posted by sammy28

hi all,

the question posed is

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

is this correct?
...
solving

$b=-7$

$a=49$

That's the same as what I got.
August 28th 2009, 01:33 AM
Defunkt

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

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

Then, you get: $x=0$ :

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

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

This gives us two possible solutions, however $a=b=0$ is obviously wrong! so we are left with $\boxed{b = -7, a = 49}$
August 28th 2009, 02:07 AM
bandedkrait

Since two quadratic expressions,

$x^2+2bx+b^2$

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

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

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

So $b=-7 , a=49$
August 28th 2009, 02:22 AM
sammy28

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

defunkt i like your logic! (Clapping)

thanks a lot
sammy