(**Mathematics 155, Section 1.4 - Miscellaneous Equations)

**Directions: Find the real solutions of each equation.**(**Note: in this section, we are specifically working on factoring and grouping along with some radical equations. I know i'm supposed to group/factor somehow but I'm confused on this one problem from this section.)

the problem is written:

**(a^2 - 6a)^2 - 2(a^2 -6a) - 35 = 0**

Ok, so, First, I factor out a^2 - 6a and group what's remaining b/c I have 4 terms and (a^2 - 6a) is common in both groups...

(a^2 - 6a) (a^2 - 6a) - 2(1) - 35 = 0 (I think this is wrong?)

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*(**Afterthought while previewing before hitting send: Should i just distribute everything and THEN factor out everything?)*

{So I started over and distributed first and came up with : (a^2 - 6a)^2 - (2a^2 + 12a) - 35 = 0

then factored out : (a^2 - 6a) (a^2 - 6a) - 2 - 2 - 35 = 0

factor out another 'a' and clean it up: a(a-6)(a-6)-39=0

a = 0 , (a-6)(a-6)-39=0

a^2 -12a +36-39 == a^2 - 12a - 3 = 0 is not factorable ... so that didn't work:

and (a-6)^2 = 39 .... and take the square root of both sides.. doesn't work either... (As you can see, I truly have worked this problem many times, in many ways.)

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(a^2 - 6a) (a^2 -6a) - 37 = 0

even if i factor out another a:

a (a-6)(a-6) - 37 = 0

so if i go with this, then i can assume that one solution is going to be

**a = 0**(so i'll just ignore that 'a' now?)

(a-6)(a-6) - 37 = 0

a^2 -12a +36 - 37 = 0

a^2 - 12a - 1 = 0

so there's no way this can work.... (unless i resort to the quadradic formula, and I don't think that's what he's looking for at this point...)

so if i go back up to my last factoring:

**(a-6)(a-6) = 37**

a = 43 ?

so my answers are 0, 43?

I still don't think I'm doing this right - and I'm really not sure where I'm going wrong or what I need to correct.

Help would be greatly appreciated.

Thank You,

Stephanie