Question #b4d9d

1 Answer
Feb 14, 2018

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# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad 1 / \alpha^2 + 1 / \beta^2 \ = \ 15. #

Explanation:

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# "We are given:" \ \ \alpha, \beta \ \ "are the zeros of the quadratic polynomial:"#

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 25x^2 + 25 x + 5 . #

# "So:" \qquad \qquad \qquad \quad \ \ \alpha, \beta \ \ "solve" \qquad 25x^2 + 25 x + 5 \ = \ 0. #

# "So:" \quad \quad \ \ alpha, \beta \ \ "also solve" \qquad \ 1/25 \cdot ( 25x^2 + 25 x + 5 ) \ = \ 1/25 \cdot 0. #

# "And:" \qquad \qquad \quad \quad \alpha, \beta \ \ "also solve" \qquad \ x^2 + x + 1/5 \ = \ 0. #

# "Thus:" \qquad \ \ \alpha, \beta \ \ "are the solutions of" \quad \ x^2 + x + 1/5 \ = \ 0. \qquad \qquad \quad (1) #

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# "Now recall that, for the quadratic equation of the form:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad x^2 + p x + q \ = \ 0, #

# \qquad \qquad \qquad "The sum of its roots is:" \qquad \qquad \qquad \ - p. #

# \qquad \qquad \qquad "The product of its roots is:" \qquad \quad + q. #

# "So, looking at (1) above, we see" \ \ p = 1 \quad "and" \quad \ q = 1/5 \ \ "for the" #
# "quadratic equation there. So we have:" #

# \qquad \alpha + \beta \ = -p = -1 \qquad "and" \qquad \alpha \beta \ = + q \ = \ 1/5. \qquad \qquad \qquad \quad \ (2) #

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# "We want the value of:" \qquad \qquad \quad \ Q \ = \ 1 / \alpha^2 + 1 / \beta^2. #

# "The idea will now be to try to rewrite" \ \ Q \ \ "as below, with an eye" #
# "toward taking advantage of the results we have in (2):" #

# \qquad \qquad \quad Q \ = \ 1 / \alpha^2 + 1 / \beta^2 \ = \ \beta^2 / { \alpha^2 \cdot \beta^2 }+ \alpha^2 / { \beta^2 \cdot \alpha^2 } #

# \qquad \qquad \qquad = \ { \alpha^2 + 2 \alpha \beta + \beta^2 - 2 \alpha \beta } / { \alpha^2 \beta^2 } \ = \ { ( \alpha + \beta )^2 - 2 ( \alpha \beta ) } / ( \alpha \beta )^2 #

# \qquad \quad "using (2):" #

# \qquad \qquad \qquad = \ { ( -1 )^2 - ( 2 \cdot 1 / 5 ) } / ( 1 / 5 )^2 \ = \ { ( -1 )^2 - 2 / 5 } / { 1 / 5^2 } \cdot 5^2 / 5^2 #

# \qquad \qquad \qquad = \ { ( -1 )^2 5^2- ( 2 / 5 ) 5^2 } / { ( 1 / 5^2 ) 5^2 } \ = \ { 25 - ( 2 / color{red}cancel{ 5 } ) 5^color{red}cancel{ 2 } } / { ( 1 / color{red}cancel{ 5^2 } ) color{red}cancel{ 5^2 } } #

# \qquad \qquad \qquad = \ { 25 - 10 } / { 1 } \ = \ 15. #

# " This is our answer. " #

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# "Summarizing:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad 1 / \alpha^2 + 1 / \beta^2 \ = \ 15. #