Question

plz help me out guys...

Answer 1

Answer:

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Answer 2

let remaining two zeroes be a and b,

then

Sum of zeroes =9/2,

2+√3 + 2-√3 +a+b=9/2,

4+a+b=9/2,

then

a+b=9/2 - 4,

then

a+b=1/2 ...............eq(1),

now
product of zeroes= -1/2,

(2+√3).(2-√3).a.b =-1/2,

[(2)²-(√3)²].ab= -1/2,

(4-3).ab= -1/2,

then

ab = -1/2 .........eq(2),


hence

(a-b)²=(a+b)²-4ab,

(a-b)²=(1/2)² - 4×-1/2,

(a-b)²=1/4 + 2,

(a-b)²=9/4,

then
(a-b)=3/2 .............eq(3),

now solve eq(1) and eq(2), we get

2a=2,

a=1,

then

1+b=1/2,

b=1/2 - 1,

b= -1/2,

therefore
other two zeroes will be 1 and -1/2