If one of the roots of x 2 -bx+c=0 , (b,c belong to rational numbers) is (7-4(3) 1/2 ) 1/2 then : (A) log b c=0 (B) b+c=5 (C) log c b=0 (D) bc=-4

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Arun · Answer

Dear student when simplified [7 – 4 sqrt(3)]^1/2 = 2 – sqrt (3) hence another root = 2 + sqrt (3) now sum of roots = b = 4 multiplication of roots = c = 4 – 3 = 1 hence log b c = 0 b + c = 4 + 1 = 5 log c b is not defined

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If one of the roots of x 2 -bx+c=0 , (b,c belong to rational numbers) is (7-4(3) 1/2 ) 1/2 then : (A) log b c=0 (B) b+c=5 (C) log c b=0 (D) bc=-4

If one of the roots of x²-bx+c=0 , (b,c belong to rational numbers) is (7-4(3)^1/2)^1/2 then :
(A) log_bc=0
(B) b+c=5
(C) log_cb=0
(D) bc=-4

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If one of the roots of x 2 -bx+c=0 , (b,c belong to rational numbers) is (7-4(3) 1/2 ) 1/2 then : (A) log b c=0 (B) b+c=5 (C) log c b=0 (D) bc=-4

If one of the roots of x2-bx+c=0 , (b,c belong to rational numbers) is (7-4(3)1/2)1/2 then :(A) logbc=0(B) b+c=5(C) logcb=0(D) bc=-4

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If one of the roots of x²-bx+c=0 , (b,c belong to rational numbers) is (7-4(3)^1/2)^1/2 then :
(A) log_bc=0
(B) b+c=5
(C) log_cb=0
(D) bc=-4