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A quadratic polynomial $p(x)$ is constructed by interpolating the data points $(0,1)$, $(1,e)$, $(2,e^2)$. If $\sqrt{e}$ is approximated by using $p(x)$ then its approximated value is.

red_trumpet
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A.MAHATA
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  • What did you try? Do you know why 3 points are required to find a quadratic polynomial? – rtybase Aug 26 '18 at 16:20
  • yes sir there are three point is given because there are three unknown in qudratic poynomial – A.MAHATA Aug 26 '18 at 16:27
  • Great, did you try to find those? It's a system of linear equations. – rtybase Aug 26 '18 at 16:28
  • first we assume the polynomial $p(x)=ax^2+bx+c$ and using the given point .We determine the coefficient but we get a absurd result which is not equal to the exact result you can help me – A.MAHATA Aug 26 '18 at 17:00
  • Add these details to the question, so that we (the entire community) can see your calculations and spot the problems. – rtybase Aug 26 '18 at 17:05
  • Hi! Welcome to Math SE! I would like you to know that to get a good response to your question you should edit it to add more details ie. what is your level of understanding and your approach etc. Else very few or none would bother to write a detailed answer for you.See-https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question. Hope you have a good time here!;-) – Soham Sep 14 '18 at 05:42

1 Answers1

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Assume $p(x)=ax^2+bx+c$ this quadratic polynomial satisfies the given points

$$p(0)=1=c$$ $$p(1)=e=a+b+c$$ $$p(2)=e^2=4a+2b+c$$

you got $a=\frac{(e-1)^2}{2}$, $b=\frac{(e-1)(3-e)}{2}$, $c=1$

thus your polynomial is $$p(x)=\frac{(e-1)^2}{2}x^2+\frac{(e-1)(3-e)}{2}x+1$$

$$\frac{(e-1)^2}{2}x^2+\frac{(e-1)(3-e)}{2}x+1=0$$ $$(e-1)^2x^2+(e-1)(3-e)x+2=0$$

Moo
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Deepesh Meena
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