Question

Find all the zero’s of 2x4− 3x3− 3x2+ 6x − 2 and its two zero’s are √2, and -√2

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

x=√2 and x = -√2  

(x -√2)=0 and (x +√2) =0  

Therefore  

(x -√2)(x +√2) =0  

(x^2 -2) = 0....................(i)  

2x4 – 3x3 – 3x2 + 6x – 2 .........(ii)  

Divide (ii) by (i)  

(2x^4 -4x^2) - (3x^3 - 6x) +(x^2 -2)  

= 2x^2(x^2 -2) -3x(x^2 -2) +1(x^2- 2)  

= (x^2 -2)(2x^2 -3x +1) = 0  

= (2x^2 -2x -x+1) = 0  

2x(x-1)-1(x-1) =0  

(2x -1)(x-1) = 0  

x = 1, 1/2 are other roots ..........................(i)  

Hence four roots are 1, 1/2, √2 and - √2 ...............Ans