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Solve: 6x2 – 15x = 0 a) x= 2 or x= 5 b) x= 2 or x = 5.5 c) x= 0 or x= 2.5 d) x= 0 or x= 2
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hint... rewrite it like: \[\LARGE 6x^2-15x+0=0 \] now use the quadratic fomula: \[\LARGE x_{1/2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] you have: \[\LARGE \underbrace{6}_{a}x^2\;\underbrace{-15}_{b}\;x\underbrace{+0}_{c}=0 \]
substitute ...
I think that it's simpler this way: \[6x^2-15x=0\]\[x(6x-15)=0\] wich is true if x = 0 or \[6x-15 = 0\rightarrow x={15\over6}=2.5\]
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