Linear equation #1

[mathwannabe]

Hello everybody

I'm done with algebra and I'm on to linear equations.

Here's the problem: I got this equation, but I need to solve it WITHOUT turning it to quadratic or higher degree equation. This is how I solved it.

1) $\displaystyle 3(z+1)^2+(z-4)^3=101+(z-3)^3$

$\displaystyle 3(z+1)^2+(z-4)^3-(z-3)^3=101$

$\displaystyle 3(z^2+2z+1)+(z-4-z+3)([z-4]^2+[z-4][z-3]+[z-3]^2)=101$

$\displaystyle 3z^2+6z+3-(z^2-8z+16)-(z^2-3z-4z+12)-(z^2-6z+9)=101$

$\displaystyle 3z^2+6z+3-z^2+8z-16-z^2+3z+4z-12-z^2+6z-9=101$

$\displaystyle 27z=135$

$\displaystyle z=\frac{135}{27}$

$\displaystyle z=5$

Now, I was wondering if anyone could maybe show me some other method for solving this equation but, I repeat, without it becoming quadratic. I tried it with squaring and cubing binomials, but I always end up with second degree variable.

I accidentally started this in the wrong section, so it would be great if moderator would move it.

 

[srmichael]

Looks good to me. But I'm curious, why the aversion of quadratic?

 

[mathwannabe]

Looks good to me. But I'm curious, why the aversion of quadratic?

It's not really an aversion, I haven't done mathematics in a long time and I'm getting myself ready for the test to get into faculty so I'm following through the high school math program. I'm currently at linear equations, and at this point in the program, high school students are oblivious to quadratics. I simply want to run through the program from the perspective of a high school student, so quadratics are a no no at this point. I hope that answers your question

 

[mathwannabe]

By the way did you see SK's elegant reduction of your expression # 8

Oh yes, I saw it, it's fantastic.