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Hiiii ppls.

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Let $a$ and $b$ be the solutions to $5x^2 - 11x + 4 = 0.$ Find $\frac{1}{a} + \frac{1}{b}.$

could someone help me with this? for a and b i get   $$a=\frac{11+\sqrt{41}}{10},\:b=\frac{11-\sqrt{41}}{10}$$   and I don't know how you put that on the bottom of a fraction... tysm in advance!!!!!! :)

Oct 20, 2020
edited by Melody  Oct 21, 2020

#1
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I plugged into Wolfram Alpha, and got 19/3.

Oct 20, 2020
#2
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Use the quadratic formula to find a, b:

x = 1/10 (11 - sqrt(41)) = a
x = 1/10 (11 + sqrt(41))= b

1 / [1/10 (11 + sqrt(41))] + 1 / [1/10 (11 - sqrt(41))]

=11 / 4

Oct 20, 2020
#3
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$$If \qquad 5x^2 - 11x + 4 = 0\\ \text{and a and b are the solutions then}\\ a=\frac{11+\sqrt{41}}{10}\qquad and \qquad b=\frac{11-\sqrt{41}}{10}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{10}{11+\sqrt{41}}+\frac{10}{11-\sqrt{41}}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{10(11-\sqrt{41})+10(11+\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\$$

I think you can finish it from there.

LaTex:

If \qquad 5x^2 - 11x + 4 = 0\\ \text{and a and b are the solutions then}\\

\frac{1}{a}+\frac{1}{b}=\frac{10}{11+\sqrt{41}}+\frac{10}{11-\sqrt{41}}\\~\\

\frac{1}{a}+\frac{1}{b}=\frac{10(11-\sqrt{41})+10(11+\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\

\frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\

\frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\

Oct 21, 2020
#4
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If the roots of the equation are a and b, then

a + b = 11/5, and ab = 4/5.

So,

$$\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{11}{5}.\frac{5}{4}=\frac{11}{4}.$$

Oct 21, 2020
#5
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You are right guest, I made the maths much more messy than necessary.   :)

Melody  Oct 21, 2020