Problem
ALG-B1-M03-P016 Equation with a Block
#16
★★★☆☆ Level 3 of 5
Solve \( (x^2-3x)^2-2(x^2-3x)-8=0 \).
Set \(u=x^2-3x\).
Let \(u=x^2-3x\). Then \(u^2-2u-8=0\), so \(u=4\) or \(u=-2\).
If \(x^2-3x=4\), then \(x^2-3x-4=0\), so \(x=4\) or \(x=-1\). If \(x^2-3x=-2\), then \(x^2-3x+2=0\), so \(x=1\) or \(x=2\).
The problem teaches recognition of a repeated block.