Take your root by Equation

Algebra Level 2

If a a and b b are roots of 2 x 2 − 3 x − 1 = 0 2x^2 - 3x - 1 = 0 , find the value of a b + b a \dfrac {a}{b} + \dfrac {b}{a} .

15 6 \dfrac{15}{6} − 12 5 \dfrac{-12}{5} 11 4 \dfrac{11}{4} − 13 2 \dfrac {-13}{2}

Anish Harsha by Anish Harsha , 20, India

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2 solutions

Anish Harsha
Aug 10, 2015

Given equation is 2 x 2 − 3 x − 1 = 0 2x^2 -3x - 1 = 0
It could be written as a x 2 − b x − c = 0 ax^2 - bx - c = 0
So, a = 2 , b = − 3 , c = − 1 a = 2, b = -3, c = -1
= a + b = − b a \dfrac {-b}{a}
= − ( − 3 ) 2 \dfrac {-(-3)}{2}
= 3 2 \dfrac {3}{2}
and a b ab = − 1 2 \dfrac{-1}{2}
= a b \dfrac{a}{b} + b a \dfrac{b}{a} = a 2 + b 2 a b \dfrac {a^2 + b^2}{ab}
= 13 4 ∗ − 2 \dfrac{13}{4} * -2
= 13 2 \dfrac{13}{2}

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Department 8
Aug 10, 2015

We have, a b + b a = a 2 + b 2 a b \frac { a }{ b } +\frac { b }{ a } \\ =\frac { { a }^{ 2 }+{ b }^{ 2 } }{ ab }

We know a b = − 1 2 ab=\frac{-1}{2} given by Vieta and a 2 + b 2 = ( a + b ) 2 − 2 a b a^{2}+b^{2}=(a+b)^{2} - 2ab and by Vieta again we find a 2 + b 2 = ( a + b ) 2 − 2 a b = 13 4 a^{2}+b^{2}=(a+b)^{2} - 2ab=\frac{13}{4} . Then:-

a b + b a = a 2 + b 2 a b = 13 4 − 1 2 = − 13 2 \frac { a }{ b } +\frac { b }{ a } \\ =\frac { { a }^{ 2 }+{ b }^{ 2 } }{ ab } =\frac{\frac{13}{4}}{\frac{-1}{2}}=\frac{-13}{2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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