Another math problem

[bethbuchall]

This is from the PSAT practice test and has us stumped. Generally, we don't have a problem with math, and we were able to get all the rest.

x^2 + 16x + a = (x + b)^2

In the equation above, a and b are constants. If the equation is true for all values of x, what is the value of a?

(Note: I wasn't sure how to format superscripts. x^2 is "x-squared")

 

[AshleyW]

a=64.

Here's a site that breaks down how to solve it:

http://www.algebra.com/algebra/home...mmutative-properties.faq.question.219013.html

 

[bethbuchall]

THANK YOU!

I can't believe that we didn't see that. You should have seen the complicated things that we were all trying to do to that problem!

 

[AshleyW]

No problem! I love math, it always winds up turning into some sick game for me to try and figure problems out, haha!

 

[damo]

The trick in that question is realizing that they say it is true for ALL values of x, so that no matter what value you put in for x, you need to get the same a and b.

Another way to do it is to let x=0, giving a=b^2. Then let x=1 and sub in a = b^2 giving 16 +b^2=2b+b^2 or b=8 and hence a = 64.

 

[pixiewings71]

No problem! I love math, it always winds up turning into some sick game for me to try and figure problems out, haha!

Ok, I know who I'm asking next time my girls have a math problem I can't solve...LOL Which is almost daily because I suck hardcore at math. LOL

 

[bethbuchall]

The trick in that question is realizing that they say it is true for ALL values of x, so that no matter what value you put in for x, you need to get the same a and b.

Another way to do it is to let x=0, giving a=b^2. Then let x=1 and sub in a = b^2 giving 16 +b^2=2b+b^2 or b=8 and hence a = 64.

Thanks! That is closer to the way that we were trying to solve it, but for some reason we never fully made the connection.

Usually I'm so good at algebra, too. I'm not sure where my mind was going.

 

[Tigger1]

This equation is also in the form of a perfect square trinomial.
The linear term has a coefficient of 16, therefore a = (16/2)^2
b = (16/2)