mathwannabe
Junior Member
- Joined
- Feb 20, 2012
- Messages
- 122
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.
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.
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