Question

1
If ß and
B are zeroes of (a2 + a)x2 + 61x + 6a. Find the values of a.
B
-​

Answer 1

Step-by-step explanation:

Band 1/B are the roots {zeros } of given

polynomial (a + a}x +61x +6a

Product of roots = constant/coefficient of

x2

B.1/B 6a/(a2+ a)

1 6a/(a + a)

a +a - 6a = 0

a - 5a =0 » a=0 and 5, a*0

a= 55

Sum of roots = - coefficient of x/

coefficient of x2

B+1/ß-61/ (a+ a) = -61/(5 +5) = -61/30

30B +30ß +61 0

Here, Discriminant = (30)-4x 30 x 61 <0

So, there is no real roots of B

Hence, a = 5 and B ER