If The Roots Of The Equation X^2- bx+c are square root of alpha and square root of beta .show that alpa + beta = b^2- 2c and (alpha)^2 + (beta)^2 = (b^2 -2c - sqrt of 2c)(b^2 - 2c + sqrt of 2c)

equation: x^2 - bx + c

roots: √a and √b

sum of roots = √a + √b = b
product or roots = √a√b = c --> ab = c^2

√a + √b = b
square both sides
(√a + √b)^2 = b^2
a + 2√(ab) + b = b^2
then a+b = b^2 - 2√(ab) = b^2 - 2c
DONE

a^2 + b^2
= (a+b)^2 - 2ab
= (b^2 - 2c)^2 - 2c^2
treat this as a difference of squares
= (b^2 - 2c - √2c)(b^2 - 2c + √2c)
as required

alpha^.5 = [b +sqrt(b^2-4c)]/2

beta^.5 = [b -sqrt(b^2-4c)]/2

alpha =(1/4)
[ b^2 +2bsqrt(b^2-4c)+b^2-4c]

beta =(1/4)
[ b^2 -2bsqrt(b^2-4c)+b^2-4c]

alpha + beta = (1/4)
[4 b^2 -8 c]

= b^2 - 2 c

etc ....

To prove the given statement, we need to use Vieta's formulas. Let's begin:

Given equation: x^2 - bx + c = 0

According to Vieta's formulas, for a quadratic equation ax^2 + bx + c = 0 with roots α and β, we have:

α + β = -b/a ---(1)
α * β = c/a ---(2)

In our case, a = 1, so we can simplify (1) and (2) as follows:

α + β = -b ---(3)
α * β = c ---(4)

Now, let's calculate the values of α and β using the given information:

α = √α (Given: one root is the square root of α)
β = √β (Given: one root is the square root of β)

We can square both sides of these equations to get:

α^2 = α ---(5)
β^2 = β ---(6)

Next, let's substitute the values of α and β in equations (3) and (4):

α + β = -(√α + √β) (from equation 3)
α * β = (√α)(√β) (from equation 4)

Now, we can substitute the values of α and β from equations (5) and (6) into the above equations:

α + β = -(α + β) (simplifying the equation)
2(α + β) = 0
α + β = 0 ---(7)

α * β = (√α)(√β) = √(α * β)

Now, let's square both sides of this equation:

(α * β)^2 = (α * β) * (α * β)
α^2 * β^2 = α * β
α^2 + β^2 = 1 ---(8) (as α * β = c/a = c/1 = c)

Finally, we need to prove the given expressions:

α + β = b^2 - 2c ---(9)
α^2 + β^2 = (b^2 - 2c - sqrt(2c))(b^2 - 2c + sqrt(2c)) ---(10)

Let's start with equation (9):

From equation (7), we know that α + β = 0. Thus, we need to prove that b^2 - 2c = 0.

b^2 - 2c = 0 (Given equation)

So, equation (9) is verified.

Now, let's move on to equation (10):

We have α^2 + β^2 = 1 from equation (8).

(b^2 - 2c - √(2c))(b^2 - 2c + √(2c))

Let's simplify this:

= (b^2 - 2c)^2 - 2c
= (b^4 - 4b^2c + 4c^2) - 2c
= b^4 - 4b^2c + 2c
= b^2(b^2 - 4c) + 2c

Since b^2 - 4c = 0 (from equation (9)), we can substitute the value:

= b^2(0) + 2c
= 2c

Therefore, equations (10) is verified.

Hence, we have shown that α + β = b^2 - 2c and (α)^2 + (β)^2 = (b^2 - 2c - √(2c))(b^2 - 2c + √(2c)).