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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.

Bobson Dugnutt
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user305570
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1 Answers1

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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} $

MathematicsStudent1122
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  • I'm having troubles with coming up with the x. Would you be able to show me how you got it? – user305570 Feb 28 '16 at 08:33
  • This is done by the quadratic formula, which you should have been taught. Otherwise, I refer you here: http://math.stackexchange.com/questions/49229/why-can-all-quadratic-equations-be-solved-by-the-quadratic-formula One word of advice: sometimes, it's difficult to determine whether or not a given quadratic equation may be factored easily. What you must do is determine whether or not the discriminant, $b^2 - 4ac$, is a perfect square. If it is, it may be possible to factor the expression nicely. – MathematicsStudent1122 Feb 28 '16 at 08:42