Question

If 2/3 and -3 are zeroes of polynomial ax^2+7x+b then find values of a and b

Answer 1

Answer:

a = -75/2 ,

b = -12

Explanation:

Let p(x) = ax²+7x+b

i) if 2/3 is a zero of p(x) , then

p(2/3) = 0

=> a(2/3)²+7(2/3)+b = 0

=> 4a/9 + 14/3+b = 0

=> (4a+42)/9 +b = 0

=> b = -(4a+42)/9 ---(1)

ii) If -3 is a zero of p(x) then

p(-3) = 0

=> a(-3)²+7(-3)+b=0

=> 9a-21+b = 0----(2)

Substitute (1) in (2) we get

=> 9a-21-(4a+42)/9=0

=> (81-189-4a-42)/9 = 0

=> -150-4a = 0

=> -4a = 150

=> a = 150/(-4)

=> a =- 75/2

Now put a = -75/2 in (1) , we get

b =- [4(-75/2)+42]/9

=> b = (-150+42)/9

=> b = -108/9

=> b = -12

Therefore,

a = -75/2 ,

b = -12

••••

Answer 2

Answer:

Hope this helps you ✌️✌️