Question

I am trying to figure out what number would make this problem a perfect squared trimomial? x^2+13x

Answer 1

well a perfect square

(x + a)^2 = x^2 + 2ax + a^2

so use the coefficient of the middle term, which is 13... divide it by 2 and then square the answer.

hope it helps

Answer 2

I guess I am just not understanding what my teacher is asking for because 13 can not be squared

Answer 3

the unit I am in is quadratic equations ugg I am frustrated

Answer 4

if you look at the perfect square

(x + a)^2 = x^2 + 2ax + a^2

13 isn't a square number... I think is what you were trying to say.

the task is to find a

so you know the middle term is 13x

so 2ax = 13x when comparing the middle terms

that means that a = 13/2

now this is a number that can be squared

\[(\frac{13}{2})^2 = \frac{13^2}{2^2}\]

and that's what you add to make a perfect square

Answer 5

well I'd expect its

\[x^2 + 13x + c\]

so the constant is half the middle coefficient squared...

as shown above

Answer 6

The question reads as " Determine the constant that should be added to the binomial so that it becomes a perfect squared trinomial.'

\[X^2+13x\]

Answer 7

ok I think i get it

Answer 8

so if you have 13/2 = a

then

(x + a)^2 = x^2 + 2ax + a^2

so

(x + 13/2)^2 = x^2 + 2(13/2)x + (13/2)^2

Answer 9

and 2(13/2)x = 13x

so its x^2 + 13x + (13/2)^2

Answer 10

YES!!! THANK YOU! Now i get it