+122 Let $a$ and $b$ be the solutions to $5x^2 - 11x + 4 = 0.$ Find 1a+1b.
could someone help me with this? for a and b i get a=11+√4110,b=11−√4110 and I don't know how you put that on the bottom of a fraction... tysm in advance!!!!!! :)
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
+118696
If5x2−11x+4=0and a and b are the solutions thena=11+√4110andb=11−√4110 1a+1b=1011+√41+1011−√41 1a+1b=10(11−√41)+10(11+√41)(11+√41)(11−√41) 1a+1b=(110−10√41)+(110+10√41)(11+√41)(11−√41) 1a+1b=(110−10√41)+(110+10√41)(11+√41)(11−√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}\\
a=\frac{11+\sqrt{41}}{10}\qquad or \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})}\\~\\
