Question

29. If a and B are the zeroes of the polynomial 6y2 – 7y + 2, find a quadratic polynomial whose zeroes are
1÷alpha and 1÷bita​

Answer 1

Answer:

polynomial=2x^2 + 7x + 6

Step-by-step explanation:

A=alpha and B=beta

6y^2 -7y + 2

6y^2 -3y -4y +2

3y(2y -1) - 2(2y-1)

(3y-2)(2y-1)

3y-2=0

3y=2

y=2/3=A=alpha

also 2y-1=0

2y=1

y=1/2=B=beta

1/(2/3)=3/2=1/alpha=alpha(a)-this is alpha for the eq to be formed

1/(1/2)=2=1/beta=beta(b)-this is beta for the eq to be formed

alpha + beta = -b/a

3/2+2=7/2=-b/a

alpha X beta = c/a

3/2*2=3/1=c/a

3*2/1*2=6/2=c/a

therefore,

a=2

b=7

c=6

polynomial=2x^2 + 7x + 6

Answer 2

Step-by-step explanation:

6y2 - 7y +2

6y2 -4y-3y +2

2y( 3y -2) -1 (+3y -2)

(3y-2) (2y -1)

alpha = 2/3 & ß= 1/2

(à+ ß) = -b/ a & àß = c/a

2/3+1/2= -b/a

7/6= -b/a

b = (-7) & a = 6 -------------(I)

àß = c/a

2/3 x 1/2 = 2/6

2/6 = c/a

c= 2 --------------------(ii)

à+ ß = -b/a

1/à + 1/ß = - a/ b

3/2 + 2/1 = -a / b

7/2 = - a/b

a = -7 & b = 2

also..

àß = c/a

1/à x 1/ß = a/c

3/2 x 2/1 = a/c

3= -7/c

c= -7/3

eqn will be....

-7y² + 2y -7/3

multiply eqn with -3

Ans : 21y² -6y + 7