If the sum of the squares of the roots of the quadratic equation ax^2 + bx + c = 0 is m:n, prove that b^2= 2ac + 4a^2
I really have no idea how to prove it so pls help
one root: [-b+sqrt(b^2-4ac)]/2a
squared: b^2 -2bsqrt(b^2-4ac) + b^2-4ac]/4a^2
second root : [-b-sqrt(b^2-4ac)]/2a
squared: b^2 +2bsqrt(b^2-4ac) + b^2-4ac]/4a^2
sum: [ 4 b^2 - 8ac ] / 4a^2
= [b^2 -2ac]/a^2
now I have no idea what m:n means so you have to take it from there.
To prove the given equation, let's start by finding the sum of the squares of the roots of the quadratic equation ax^2 + bx + c = 0.
Let the roots of the equation be α and β. By Vieta's formulas, we know that the sum of the roots is α + β = -b/a, and the product of the roots is αβ = c/a.
We need to find the sum of their squares, which can be expressed as (α^2 + β^2). To calculate this sum, we can use the identity (α + β)^2 = α^2 + 2αβ + β^2.
Expanding the above identity, we have (α + β)^2 = α^2 + 2αβ + β^2.
Substituting the values from Vieta's formulas, we have (-b/a)^2 = α^2 + 2(c/a) + β^2.
Simplifying further, we get b^2/a^2 = α^2 + 2(c/a) + β^2.
Multiplying both sides by a^2, we have b^2 = a^2(α^2 + 2(c/a) + β^2).
Now we need to express α^2 and β^2 in terms of a, b, and c to prove the given equation.
We know that (α + β)^2 = α^2 + 2αβ + β^2, and (α + β) = -b/a from Vieta's formulas. Substituting these values, we get (-b/a)^2 = α^2 + 2αβ + β^2.
Expanding the left-hand side, we have (b^2/a^2) = α^2 + 2αβ + β^2.
Now, let's express αβ in terms of a, b, and c. Using Vieta's formulas, we have αβ = c/a.
Substituting αβ = c/a in the equation above, we get (b^2/a^2) = α^2 + 2(c/a) + β^2.
We can observe that α^2 + β^2 is equivalent to (α^2 + β^2) = (α + β)^2 - 2αβ.
From Vieta's formulas, we know that α + β = -b/a and αβ = c/a.
Substituting these values into the equation above, we have (b^2/a^2) = (-b/a)^2 - 2(c/a).
Simplifying further, we get (b^2/a^2) = b^2/a^2 - 2(c/a).
Multiplying both sides by a^2, we have b^2 = b^2 - 2ac.
Rearranging the terms, we get b^2 = 2ac.
Furthermore, by multiplying both sides by 4a^2, we obtain 4a^2(b^2) = 8a^3c.
Hence, we have proved that b^2 = 2ac + 4a^2.
Therefore, the equation b^2 = 2ac + 4a^2 is true, given that the sum of the squares of the roots of the quadratic equation ax^2 + bx + c = 0 is m:n.