if the roots of the equation ax2+bx+c=0 be four times the other root, show that 4b2 = 25ac.

Comparing the given equation with ax² + bx + c = 0, we get,

a= 2, b = –9 and c = 35

Then,

b² – 4ac = (–9)² – 4.2.35

= 81 – 280

= –199

Here, b² - 4ac < 0

So, the roots are imaginary and unequal.

Solⁿ:

Let α be the common root of the given equations:

α2 + pα + q = 0 …(i)

and α2 + p’α + q’ = 0 …(ii)

Solving (i) and (ii) by cross multiplication,

1 p q 1

1 p’ q’ 1

Or, α²/(pq′p′q) = α/(qq′) = 1/(p′p)

Then,

α²/(pq′p′q) = α/(qq′) or, α/(qq′) = 1/(p′p)

α = (pq′p′q)/(qq′) or, α = (qq′)/(p′p)

So, α be either (pq′p′q)/(qq′) or. (qq′)/(p′p)

Soln:

Let the roots be pα and qα.

Then, pα + qα = –n/l and pα.qα = n/l

Now,

(pα+qα)/pα.qα = –(n/l)/(n/l)

Or, α(p+q)/α(pq) = –(n/l)

Or, p/pq + q/pq = –(n/l)

Or, p²/pq + q²/pq+ n/l = 0

So, p/q + q/p + n/l = 0