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## 6. α, β are roots of $y^2$ – 2y – 7 = 0 find, (1) $a^2$ + $ß^2$ (2) $a^3$ + $ß^3$

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Sol.     $y^2$ – 2y – 7 = 0 comparing the equation
with a$x^2$ + b$x$ + c = 0, we get

a = 1   b = –2   c = –7

∴         α   +   β   =   $-{b\over a}$    =  ${-(-2)\over 1}$  =  2

∴             αβ       =     $c\over a$     =   $-7\over 1$     = -7

(1)    $a^2$ + $ß^2$

 By using formula    $(a+b)^2=a^2+2ab +b^2$  ∴   $(a+b)^2-2ab=a^2+b^2$

∴    $a^2$ + $ß^2$  =    $(a + ß)^2$ – 2αβ
=    $2^2$ – 2 × (–7)
=    4 + 14
∴    $a^2$ + $ß^2$  =   18

(2)  $a^3$ + $ß^3$

 By using formula $(a+b)^3=a^3+ 3ab(a + b) +b^3$ ∴ $(a+b)^3- 3ab(a + b)=a^3 +b^3$

$a^3$ + $ß^3$  =  $(a + ß)^3$  –  3αβ (α + β)
=  $(2)^3$ – 3(–7) × 2
=   8 + 42
∴      $a^3$ + $ß^3$  =   50

uÉaÉïxÉqÉÏMüUhÉ August 10 , 2018 0 Comments 133 views