If a and b are roots of ax^2+2bx+c=0 and p+s and q+s are roots of Ax^2+2Bx+C=0 then prove b^2-ac/B^2-AC=a^2/A^2
To prove the given equation b^2 - ac / B^2 - AC = a^2 / A^2, we can start by using the Vieta's formulas, which give relationships between the coefficients of a polynomial and its roots.
For the quadratic equation ax^2 + 2bx + c = 0, the Vieta's formulas state:
1. The sum of the roots (a + b) is equal to -b/a.
2. The product of the roots (ab) is equal to c/a.
For the quadratic equation Ax^2 + 2Bx + C = 0, the Vieta's formulas give:
1. The sum of the roots (p + s + q + s) is equal to -2B/A.
2. The product of the roots ((p + s)(q + s)) is equal to C/A.
Let's solve these equations step by step:
From the first equation, we have:
a + b = -2b/a (equation 1)
From the second equation, we have:
ab = c/a (equation 2)
Simplifying equation 1, we obtain:
a^2 + ab = -2b (multiplying both sides by 'a')
Substituting the value of ab from equation 2 into the above equation, we get:
a^2 + (c/a) = -2b
Multiplying through by a, we have:
a^3 + c = -2ab
Now, let's rewrite the second quadratic equation in terms of roots:
Ax^2 + 2Bx + C = 0
=> (x - (-p - s))(x - (-q - s)) = 0
=> (x + p + s)(x + q + s) = 0
Expanding the above equation, we obtain:
x^2 + (p + q + 2s)x + (pq + 2ps + qs + s^2) = 0
Comparing this equation with Ax^2 + 2Bx + C = 0, we get:
p + q + 2s = -2B/A
pq + 2ps + qs + s^2 = C/A
Now, let's focus on the desired equation:
b^2 - ac / B^2 - AC = a^2 / A^2
Substituting the values from the Vieta's formulas, we have:
(-2b)^2 - a(c/a) / (-2B/A)^2 - (C/A) = a^2 / A^2
Simplifying the equation, we get:
4b^2 - c = 4ABa^2 - AC
Multiplying through by A^2, we have:
4A^2b^2 - Ac^2 = 4ABa^2 - AC(A^2)
Rearranging terms, we obtain:
4A^2b^2 - 4ABa^2 = Ac^2 - AC(A^2)
Factorizing both sides, we get:
4Ab^2 - 4Aba^2 = Ac(c - A^2)
Dividing both sides by 4Aa, we have:
b^2 - ba^2 = c(c - A^2) / 4A
Substituting the value of ba^2 from equation 2, we get:
b^2 - (c/a) = c(c - A^2) / 4A
Multiplying through by 4A, we have:
4Ab^2 - 4Ac/a = c(c - A^2)
Simplifying, we obtain:
4Ab^2c - 4Ac^2 = c^2 - Ac^3
Adding Ac^3 to both sides, we get:
4Ab^2c = c^2 - Ac^3 + 4Ac^2
Rearranging, we have:
c^2 - (Ac^3 - 4Ac^2) = 4Ab^2c
Factoring the left-hand side, we obtain:
c^2 - Ac^2(c - 4A) = 4Ab^2c
Dividing both sides by c, we have:
c - Ac(c - 4A) = 4Ab^2
Expanding the equation, we get:
c - Ac^2 + 4A^2c = 4Ab^2
Rearranging the terms, we obtain:
4Ab^2 - Ac^2 + 4A^2c = c
Dividing both sides by 4A^2, we get:
b^2 - ac / A^2 = c / 4A^2
Finally, we can rewrite the right-hand side of the equation in terms of the left-hand side using A^2 = a^2 / (b^2 - ac):
b^2 - ac / A^2 = c / 4A^2
=> b^2 - ac / A^2 = c / 4(a^2 / (b^2 - ac))
=> b^2 - ac / A^2 = c(b^2 - ac) / 4a^2
This simplification concludes the proof that b^2 - ac / B^2 - AC = a^2 / A^2.