Question
I. 3x6- 19x3+16=0 II.
9y4- 27y2+20=0 In the following questions two equations numbered I and II are given. You have to solve both the equations. Give answer if;Solution
Let x3=b 3b2-19b+16=0 3b2-16b-3b+16=0 Pairs are = -16, -3 Values after changing sign = 3, 16 & after dividing by 3, final values of b = 3/3 , 16/3 x3=1 ,16/3 x=1 , -1 ,∛16/∛3 II) Let y2=a 9a2-27a+20=0 Pairs are -15,-12 Values after changing sign = 15, 12 & after dividing by 9, final values of a = 15/9 , 12/9 = 5/3 , 4/3 y²= 4/3 ,5/3 y= 2/√3, -2/√3, √5/√3 ,-5/√3 Hence, the relationship cannot be established
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