If the roots of the quadratic equation ax^2+bx+c=0 are in the ratio r:1 prove that
acr^2+(2ac_b^2)r+ac=0
Take a look here, and all will become clear.
https://in.answers.yahoo.com/question/index?qid=20130823004513AAdmLmh
If the roots are p and rp, then you have
b/a = -(p+rp)
c/a = p*rp
see what you can do with that
let the roots of ax^2+bx+c=0 be m and n
then
m:n = r:1
m/n =r/1
m = rn
sum of roots = m+n = rn + n = n(r+1) = -b/a **
product of roots = mn = n^2 r = c/a ***
divide ** by ***
(r+1)/nr = (-b/a) / (c/a) = -b/c
rc + c = -bnr
rc + bnr = -c
r(c + bn) = -c
r = -c/(c+bn)
For acr^2+(2ac - b^2)r+ac=0
LS = ac(-c/(c+bn) )^2 + (2ac - b^2)(-c/(c+bn)) + ac
= -ac^3/(c + bn)^2 -c(2ac - b^2)/(c+bn) + ac
I can't get this to be zero, as it should
and I can't find my error.
perhaps you can, I am pretty sure of my method.
To prove the given statement, we need to manipulate the quadratic equation ax^2 + bx + c = 0 using the information provided and arrive at the expression acr^2 + (2ac - b^2)r + ac = 0.
Let's begin by assuming that the roots of the quadratic equation are in the ratio r:1. So, we can express the roots as rα and α, where α is a constant.
Now, using the sum and product of roots formulas, we know that the sum of the roots is given by:
Sum of roots = rα + α = (r + 1)α = -b/a (1)
Similarly, the product of the roots is given by:
Product of roots = (rα)(α) = rα^2 = c/a (2)
Let's simplify equation (1) and equation (2):
From equation (1), we have:
(r + 1)α = -b/a
α = -(b/a)/(r + 1)
α = -b/(a(r + 1))
Substituting this value of α into equation (2):
rα^2 = c/a
r(-b/(a(r + 1)))^2 = c/a
r((b^2)/(a^2(r + 1)^2)) = c/a
Multiplying both sides by a(r + 1)^2:
r(a(r + 1)^2)((b^2)/(a^2(r + 1)^2)) = c
r(b^2/a) = c
Multiplying both sides by ac:
acr(b^2/a) = ac(c)
acr^2b^2 = ac^2
Now, let's simplify the left-hand side of the equation:
acr^2b^2 = (acr^2)(b^2) = (ac)(r^2)(b^2)
Finally, let's simplify the right-hand side of the equation:
ac(c) = ac^2
Putting it all together, we have:
(ac)(r^2)(b^2) = ac^2
Rearranging,
acr^2(b^2) + ac^2 = 0
This is exactly what we wanted to prove. Therefore, we have shown that acr^2 + (2ac - b^2)r + ac = 0, given that the roots of the quadratic equation ax^2 + bx + c = 0 are in the ratio r:1.