# What are the values of a and c in the trinomial square ax2(the two is an exponent)+12x+c if a doesnt equal 1 and c is greater than a?

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### 1 Answer

Given the perfect square trinomial `ax^2+12x+c` , find the values for a and c if `cgta, a!=1` :

Since this is a perfect square trinomial we have:

`(sqrt(a)x+sqrt(c))^2=ax^2+12x+c`

Expand the left side and equate like terms:

`ax^2+2sqrt(ac)x+c=ax^2+12x+c`

`2sqrt(ac)=12`

`sqrt(ac)=6`

`ac=36`

Now we have the restrictions `a!=1,c>a` . Restricting to integral solutions we get the following possible solutions:

a=2,c=18

a=3,c=12

a=4,c=9

**The only solution with integral coefficients is a=4,c=9 yielding:**

`(2x+3)^2=4x^2+12x+9`

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If we are not restricted to integral coefficients for the binomial we get:

a=2,c=18 ==> `(sqrt(2)x+sqrt(18))^2=2x^2+12x+18` and

a=3,c=12 ==> `(sqrt(3)x+sqrt(12))^2=3x^2+12x+12`

There are an infinite number of solutions if a and c are not forced to be integral.