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Let a and b be the solutions to 5x^2 - 11x + 4 = 0 Find: 1/a^2 + 1/b^2.

 Apr 15, 2022
 #1
avatar+326 
+2

Using the quad formula:

 

\(\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\cdot \:5\cdot \:4}}{2\cdot \:5}\)

 

\(\frac{\left(11\right)\pm \sqrt{41}}{10}\)

 

Substituting this into \(\frac{1}{a}^2\:+\:\frac{1}{b}^2.\)

 

\(\frac{1}{\frac{\left(11\right)\pm \:\sqrt{41}}{10}}^2\:+\:\frac{1}{\frac{\left(11\right)\pm \:\sqrt{41}}{10}}^2\)

 

\(\frac{81\pm11\sqrt{41}}{16}\)

 

or 

 

approx. 9.46464646464 and 0.66035

 

-Vinculum

 

smileysmileysmiley

 Apr 15, 2022
 #2
avatar+122390 
+1

Note

 

1/a^2   +   1/b^2  =     [ a^2 + b^2 ] /  [ ab]^2

 

5x^2   -11x + 4   =  0

 

By Vieta

 

Sum of the roots =  11/5

a + b = 11/5       square both sides

a^2 + 2ab  + b^2   =  121/ 25 

a^2 + b^2  =  121/25 - 2ab      (1)

 

And product of the  roots  =  4/5

So

ab = 4/5    ⇒  [ab]^2  = 16/25

2ab = 8/5

So (1)  becomes a^2 + b^2 =  121/25 - 8/5  =   121/25 - 40 /25  =   81/25

 

So

 

1/a^2  + 1/b^2   =     [ a^2 + b^2]  / [  ab]^2   =   [ 81/25] / [ 16/25]  =   81/16

 

 

cool cool cool

 Apr 16, 2022

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