Question

if -3 , -4 , 4 are the roots of x^3-3x^2+ax+48=0 find the value of a​

Answer 1

Answer:

Given x = - 3 ,-4 ,4.

x³-3x²+ax+48 =0

Substituting the value of x as - 3,

(-3)³-3(-3)²+a(-3)+48 =0

-27 - 27- 3a +48 =0

3a = 48 - 54.

3a = - 6

a = - 2

Now Substituting the value of x as - 4,

(-4)³-3(-4)² + a(-4)+48 =0

-64 - 48 - 4a +48 =0

4a= 64

a = 16.

Then Substituting the value of x as 4,

4³-3(4)²+ 4a +48 =0

64 - 48 +4a +48 = 0

4a = - 64

a = - 16.

Therefore the three values of a are - 2, 16 and - 16.

Hope it helps you.