Find the exact real root for the equation $10x^3-12x^2-6x-1=0$.
The given cubic cannot be factored easily and beautifully, so the method of factoring the given polynomial is out of the question.
I hear you, the next best approach might be to try out the substitution method, with the hope that after the substitution, we have less variable terms in our newly set equation. But there seems no suitable substitution is available so to make simpler the given equation.
Showing posts with label algebraically manipulate. Show all posts
Showing posts with label algebraically manipulate. Show all posts
Show that [MATH]16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17[/MATH].
Show that [MATH]16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17[/MATH].
On my previous post (Show that [MATH]16<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{80}}<17[/MATH]), we used the trapezoid rule to prove the upper bound for the given inequality, i.e. [MATH]16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}[/MATH]. We're then not supposed to use the same method to prove the lower bound simply because the function $y=\dfrac{1}{\sqrt{x}}$ is concave up. No matter how we manipulated that concept, we will only end up with proving the target sum [MATH]\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}[/MATH] will be greater than some quantity, not less than. This works against to what we are looking to prove, that is, [MATH]\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17[/MATH].
On my previous post (Show that [MATH]16<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{80}}<17[/MATH]), we used the trapezoid rule to prove the upper bound for the given inequality, i.e. [MATH]16<\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}[/MATH]. We're then not supposed to use the same method to prove the lower bound simply because the function $y=\dfrac{1}{\sqrt{x}}$ is concave up. No matter how we manipulated that concept, we will only end up with proving the target sum [MATH]\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}[/MATH] will be greater than some quantity, not less than. This works against to what we are looking to prove, that is, [MATH]\sum_{k=1}^{80}\dfrac{1}{\sqrt{k}}<17[/MATH].
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