Completing the square

H

homeschool girl

Junior Member
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Feb 6, 2020
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The quadratic
$x^2+(2.6)x+3.6$
can be written in the form
$(x+b)^2+c$
, where
$b$
and
$c$
are constants. What is
$b+c$
(as a decimal)?
 
I'll walk you through the problem. Let's begin by writing:

[MATH]y=x^2+2.6x+3.6[/MATH]
We want to group together the two terms containing \(x\):

[MATH]y=(x^2+2.6x)+3.6[/MATH]
Now, we look at the coefficient of the linear term (the term with \(x\) to an implied power of 1, the term \(2.6x\)). we want to divide the coefficient \(2.6\) by 2 to get \(1.3\) and then we want to square the result to get \(1.69\). So, we add this value inside the brackets, but, we have to subtract it on the outside so that in effect we are adding zero to the expression, so that its value is not changed.

[MATH]y=(x^2+2.6x+1.69)+3.6-1.69[/MATH]
Now, we can write this as:

[MATH]y=(x^2+2\cdot1.3x+1.3^2)+1.91[/MATH]
Notice the expression within the brackets is now a square:

[MATH]y=(x+1.3)^2+1.91[/MATH]
Do you follow me so far?
 
We have written the given expression as:

[MATH]y=(x+1.3)^2+1.91[/MATH]
This is the form:

[MATH]y=(x+b)^2+c[/MATH]
So, you need to be able to identify \(b\) and \(c\) in the first equation above...
 
so on this site it tries to explain this but I need simpler or more detailed help. Can you explain how this works from start to finish with everything written out?
the sample problem freemathhelp.com gives is:
4x2+4y2-24x+32y-4=0
 
Look at (x+a)2.
(x+a)2= x2 + 2ax + a2

Now how can we go from 2a (the coefficient of the x term) to a2 (the constant)?

One way is to divide 2a by 2 (to get a) and then square the result (which is a^2)

So whenever you something in the form of x2 + rx + s where (r/2)2 = s then you have a complete square, namely (x+r/2)2

If you want further help, possibly with the problem you post, can you please make a separate post. Thanks.
 
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