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# plz help

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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
+579
+3

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

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

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

Apr 15, 2022
#2
+128794
+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

Apr 16, 2022