Question
If (x + a) is a common factor of x2 β
(2+β2)x + 2β2 and x2 - 2β2x + 2, then the value of β(5β2 β a)3β2 is :Solution
x2 β (2+β2)x + 2β2 = 0 ------ (i) x2 - 2β2x + 2 = 0 ------ (ii) Now, x + a = 0 => x = -a ------ (iii) Substituting in (i) and (ii), => x2 - (2 + β 2) x + 2β2 = 0 => a2 - ( 2 + β 2 ) (-a) + 2β 2 = 0 => a2 + ( 2 + β 2 ) a + 2β 2 = 0 => x2 - 2β 2 x + 2 = 0 => a2 + 2β 2 a + 2 = 0 Now, => a2 + ( 2 + β 2 ) a + 2β 2 = a2 + 2β 2 a + 2 2a + β 2 a + 2β 2 = 2β 2 a + 2 2a + β 2a - 2β 2 a = 2 - 2β 2 2a - β 2a = 2 - 2β 2 a ( 2 - β 2 ) =Β 2 - 2β 2 a = ( 2 - 2β 2 ) / ( 2 - β 2 ) Substituting the value of a in the equation: β ( 5β 2 - a ) 3β 2 = β ( 5β 2 - [ ( 2 - 2β 2 ) / ( 2 - β 2 ) ] Γ 3β 2 = β [ ( 5β 2 Γ ( 2 - β 2 ) ) - ( 2 - 2β 2 )Β /Β ( 2 - β 2 ) ] Γ 3β 2 = β [ ( 10β 2 - 10 - 2 + 2β 2 ) / ( 2 - β 2 ) ] Γ 3β 2 = β [ ( 12β 2 - 12 ) / ( 2 - β 2 ) ] Γ 3β 2 = β [ ( (36 Γ 2) - ( 36 Γ β 2 ) ) / ( 2 - β 2 ) ] = β ( 72 - 36β 2) / ( 2 - β 2 ) = β 36 ( 2 - β 2 ) / ( 2 - β 2 ) = β 36 = 6 β΄ β ( 5β 2 - a ) 3β 2 = 6
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