Finding the value of y=b^2(3a^2+4ab+2b^2) if a^2(2a^2+4ab+3b^2)=3 and ф and b are distinct zeros of x^3−2x+c

Demarion Ortega

Demarion Ortega

Answered question

2022-11-04

Finding the value of y = b 2 ( 3 a 2 + 4 a b + 2 b 2 ) if a 2 ( 2 a 2 + 4 a b + 3 b 2 ) = 3 and ф and b are distinct zeros of x 3 2 x + c

Answer & Explanation

dobradisamgn

dobradisamgn

Beginner2022-11-05Added 17 answers

Since a and b are the solutions of x 3 2 x + c = 0 we have,
a 3 2 a + c = 0
and
b 3 2 b + c = 0
Subtracting the above two equations, we get,
a 3 b 3 2 ( a b ) = 0
( a b ) ( a 2 + a b + b 2 2 ) = 0
a 2 + a b + b 2 = 2 (Since a and b are distinct)
Now,
a 2 + a b + b 2 = 2
a 2 ( 2 a 2 + 4 a b + 3 b 2 ) = 3
b 2 ( 3 a 2 + 4 a b + 2 b 2 ) = y
Squaring the first equation, we get,
a 4 + b 4 + 3 a 2 b 2 + 2 a 3 b + 2 a b 3 = 4
2 a 4 + 2 b 4 + 6 a 2 b 2 + 4 a 3 b + 4 a b 3 = 8
Expanding and adding the other two equations, we get,
y + 3 = 2 a 4 + 2 b 4 + 6 a 2 b 2 + 4 a 3 b + 4 a b 3
y + 3 = 8
y = 5

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