Question

what is the value of c so that x^2+13x+c is a perfect square trinomial?

Answer 1

c is (13/2)^2

Answer 2

(a+b)^2=a^2+2ab+b^2

Answer 3

g
x^2+13x+c=0
[x^2+13x+42.25]+c-42.25=0
^This step is where we will use the coefficient of x^2 (1) and divide by two and multiply it by the coefficient of the second number then square this number; (1/2)(13)=(6.5)^2=42.25 Now we need to Add and Subtract this number from inside and outside the brackets.

Divide the un-squared X by two and this will give us a simple square

Now we can break this down into a simple square [Remember 13/2 is 6.5]
(x-6.5)(x-6.5) +c -42.25=0
let's move the rest of the numbers to the other side
(x-6.5)^2=-c+42.25
Let's take the Square root of Both sides Let's call these brackets the square root /{} and when we take the square root of a number in this case we will need to either + or - this which will give us two variables
/{(x-6.5)^2}=+or-/{-c+42.25}
we also have to get rid of the -6.5 on the left side
X1=+/{-c+42.25}+6.5
AND
X2=-/{-c+42.25)+6.5

You will get a value for c and you can get a number for X1 and X2 and you plug each one in the the original equation and you will solve it.