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Grade: 11

                        

`a`, `b` be roots of x^2-1154x+1=0, then find the value of "(a)^1/4+(b)^1/4".

6 years ago

Answers : (2)

Pushkar Aditya
71 Points
							According to the equation given, a+b=1154 and ab=1.
let a^1/4+b^1/4= k
k^2= a^1/2+b^1/2+2(ab)^1/4
Since ab=1
k^2= a^1/2+b^1/2+2 
root A+ root b= k^2-2........(This value is important)
k^4= a+b+4+2(ab)^1/2+4 root A+ 4 root B
k^4= 1154+2+4+4(k^2-2)
Final equation is easy to solve and simplified
k^4-4 k^2-1152; 
put k^2=x and solve the equation to get x=36 or k^2=36 so k is our answer which is equal to 6.

I recommend to copy down the whole solution to understand properly
						
6 years ago
Apptica Corporations
37 Points
							According to the equation given, a+b=1154 and ab=1. 
let a^1/4+b^1/4= k 
k^2= a^1/2+b^1/2+2(ab)^1/4 Since ab=1 
k^2= a^1/2+b^1/2+2 
root A+ root b= k^2-2........(This value is important)
 
Now calculate k^4 and we get

k^4=a+b+4+2(ab)^1/2+4(root A + root B )
replace the root A+ root B value to form a biquadratic equation of k

k^4-4k^2-1152---> This equation is very easy to solve

let k^2=x and solve for x to get x=36 or k^2= 36 or k=6

Hence we get k=6 that is our answer I guess
						
6 years ago
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