Question

Completing the square. Find the value of n such that each expression is a perfect square trinomial.

x^-6x+n

Answer 1

@ganeshie8

Answer 2

@whpalmer4

Answer 3

To complete the square, the constant term needs to be the square of one half of the coefficient of the \(x\) term.

Answer 4

i dont understant please help me what is the cnstant term?

Answer 5

The constant term is the one that doesn't have the variable...

\[x^2+3x+5\]the constant term here is 5

Answer 6

howd u get 3>?

Answer 7

That's an example...

Answer 8

ok so in the question 6 isnt a square

Answer 9

You didn't read what I wrote quite correctly. Take the coefficient of the \(x\) term. Divide in half. Square that number. That is the number you need as the constant term to form a perfect square.

Answer 10

Let's work backwards:

a perfect square can be written as \[(x+a)^2 = (x+a)(x+a) = x^2 +2ax + a^2\]

Now, you have
\[x^2-6x+n\]\[x^2+2ax+a^2\]

for those to be equal, \[-6x=2ax\]and\[n=a^2\]

Answer 11

If we solve the first for \(a\):
\[-6x=2ax\]\[-6=2a\]\[-3=a\]
That's half of the coefficient of the \(x\) term, right?

Now we square it: \((-3)^2 = 9\)

\[(x-3)^2 = x^2 -6x + 9\]

Answer 12

\(n=9\) makes \(x^2-6x+n\) a perfect square.

Answer 13

ohh ok i get it now...

Answer 14

Now, if you are completing the square in the context of an equation, you need to be careful that you don't break the equality!

For example:

\[x^2 + 4x = 0\]Hopefully, you can easily see that we need to add \(4\) to the left side of the equation to complete the square and rewrite it as \((x+2)^2\). However, we can't just add \(4\) to one side of the equation unless we either add \(4\) to the other side of the equation, or we subtract \(4\) somewhere else on the same side where we added it. The two are of course equivalent, and which you do is just a matter of convenience.

\[(x^2+4x + 4) - 4 = 0\]\[(x+2)^2-4 = 0\]
there I added and subtracted on the same side, and \(+4 - 4 = 0\) so everything is fine.

I could also have done this:
\[x^2 + 4x + 4 = 0 + 4\]\[(x+2)^2 = 4\]
I think you should be able to see that is equivalent to the other, right?

Answer 15

hmm yeah how about if its like thsi 2x^2-8x+n

Answer 16

Well, there you need to factor out that leading 2.

\[2x^2-8x+n\] is the same as \(2(x^2-4x+n/2)\) right?
Complete the square on the stuff inside the parentheses.

Or, if you like you can do it like this:

\[ax^2+bx+c\]take half of \((b/a)\), square that, and that's your new constant term.

Answer 17

so it would be +16?

Answer 18

@whpalmer4

Answer 19

Sorry, I didn't explain that quite correctly: you also have to divide that squared value by the value of \(a\)....

\[2(x-2)^2 = 2(x^2-4x+4) = 2x^2-8x+8\]

Answer 20

and then?

Answer 21

So, the correct value was +8, not +16...

Answer 22

ok thanks can u check out teh other problem i posted? please