The question is : $2x^2-3x-1 \leq 0$
I've got the answers but I keep getting them wrong. If someone could write out the steps it would be greatly appreciated.
2026-04-13 19:14:17.1776107657
Quadratic inequalities 1
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The quadratic polynomial $2x^2 - 3x - 1$ has zeroes at $$x=\frac {3±\sqrt{17}}{4}$$
and is concave up. Hence $2x^2 - 3x - 1 \leq 0 \Longleftrightarrow\frac {3-\sqrt{17}}{4} \leq x \leq \frac {3+\sqrt{17}}{4} $