Answers

2016-04-25T10:12:50+05:30
Let the roots of the given equation be 'a' &'b' .
a+b=-(-7)/6=7/6. .....(1)
ab=2/6=1/3. .....(2)
Now,the required equation has roots: 1/a & 1/b.
sum of roots,S=(1/a)+(1/b)
=(a+b)/ab
=(7/6)÷(1/3). {from 1 & 2}
S =7/2. ...(3)
product of roots,P=(1/a)×(1/b)
=1/(ab)
=1÷(1/3)
P =3. ...(4)
The required equation will be of the form:
y^2 -Sy+P = y^2-(7/2)y+3=O [from 3 & 4]
=2y^2-7y+6=0
Hence,the required equation is:
2y^2 - 7y + 6 = 0
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2016-04-25T15:05:11+05:30
6y²-7y+2
6y²-4y-3y+2
2y(3y-2) - 1(3y-2)
(3y-2)(2y-1)

Zeroes = 2/3 or 1/2

If alpha = 2/3 and beta = 1/2

then 1/alpha = 3/2 and 1/beta = 2

Now the quadratic polynomial is 

x²-(sum of roots)x + (product of the roots) 

Sum of the roots = 3/2 + 2 = 3.5
product of the roots = 3/2 * 2 = 3

The quadratic is 

x² - 3.5x + 3
0