Question

If the roots of the equation x^3-ax^2+bx-c=0 are in the ration 2:4:5 and c=320 then value of b is

Answer 1

Answer:

Mark it as brainliest

Yes

Answer 2

The value of b is 152.

If the roots of the equation x³ - ax² + bx - c = 0 are in the ratio 2 : 4 : 5 and c = 320.

We have to find the value of b.

Let proportionality constant is k.

then roots are 2k, 4k and 5k.

sum of products of two consecutive roots = coefficient of x/coefficient of x³

⇒2k × 4k + 4k × 5k + 5k × 2k = b/1

⇒8k² + 20k² + 10k² = b

⇒38k² = b ...(2)

product of all roots = - constant/coefficient of x³

⇒2k × 4k × 5k = -(-c)/1 = c = 320

[ ∵ given c = 320 ]

⇒40k³ = 320

⇒k = 2

now, b = 38k² = 38 × 2² = 38 × 4 = 152

Therefore the value of b is 152.