Question

Solve (x^3) -(3x^2) -(16x) +48=0, given that the sum of two roots is zero



can you plz answer fast​

Answer 1

Step-by-step explanation:

Let the roots be a , b and c

such that, a + b= 0.....(1)

Now , a + b+c = -(-3)/1= 3

→ c= 3

Also , abc = -48/1

→ ab = -16 ........(2)

Solving (1) and (2),

We get

(a-b) = 8 or , -8 ......(3)

From (1) and (3)

a= 4 and ,b= -4

OR ,

a= -4 and , b = 4

Note :- (a-b)^2 = (a+b)^2-4ab

Use this formula to find (a-b)