Question

if aby are the zeros of the polynomial f(x)=ax^3 +bx^2 + cx +d then 1/a + 1/b + 1/y=?

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

if a,b and y are the zeros of the polynomial f(x)=ax^3 +bx^2 + cx +d then 1/a + 1/b + 1/y = ?

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

For a cubic equation ax³+ bx² +cx+ d = 0 let p,q and r be its roots, then the following holds :-

• p + q + r = (-b/a)
• pq + qr + pr = c/a
• p * q * r = (-d/a) .

Sᴏʟᴜᴛɪᴏɴ :-

from above told formula , when roots are a, b & y, we get :-

→ (ab + by + ay) = c/a -------- Equation (1)

→ a * b * y = (-d/a) ------------ Equation (2)

Now,

→ 1/a + 1/b + 1/y

Taking LCM ,

→ (by + ay + ab) / (aby)

Or,

→ (ab + by + ay) / aby

Putting values of Equation (1) & (2) Now, we get,

→ (c/a) / (-d/a)

→ (c/a) * (-a/d)

→ (-c/d) (Ans.)

Answer 2

GIVEN:

• a , b & y are the roots of the polynomial f(x) = ax³ + bx² + cx + d

TO FIND:

• 1/a + 1/b + 1/y = ?

SOLUTION:

Taking a cubic equation ax³ + bx² + cx + d = 0 . Having roots α , β & γ

Conditions:

♦ α + β + γ = - b/a

♦ αβ + βγ + αγ = c/a

♦ α × β × γ = - d/a

Finding 1/a + 1/b + 1/y

Using above conditions

Taking LCM

♦ ab + by + ay/aby

Simplifying using above conditions

♦ (c/a)/(-d/a)

♦ - c/d

ab + by + ay/aby = - c/d

.°. Hence, 1/a + 1/b + 1/c = - c/d