The sum of the roots of the equation x ² + px + q = 0 is equal to the sum of their squares, then

ab/(a + b) = [p/(p + q)] x [q/(p - q)] / [p/(p + q) + q/(p - q)]
= pq/p² - pq + pq + q²
= pq/(p² + q²)

Here, x² = 6 + ?6 + ?6 + ?6 + .... ? ,
So, x² = 6 + ?x²
? x² = 6 + x
? x² - x - 6 = 0
? x² + 2x - 3x - 6 = 0
? x(x + 2) - 3(x + 2) = 0
? (x - 3) (x + 2) = 0
? x = 3

Given equation is
x² - 11x + 24 = 0
Then, required equation is
(x - 2)² - 11(x - 2) + 24 = 0
? x² - 4x + 4 - 11x + 22 + 24 = 0
? x² - 15x + 50 = 0

Given, ? and ? are the roots of the equation ax² + bx + c = 0
? Sum of two roots = -b/a
? + ? = -b/a
and product of two roots = c/a
? ? = c/a
? ?²/? + ?²/? = ( ?³ + ?³) / ? ?
= [(? + ?) ³ - 3 ? ? (? + ?)] / ??
[? (a + b)³ = a³ + b³ + 3ab(a + b) ]
= (-b/a)³ - [3c/a(-b/a)/c/a = -b³/a³ + 3bc/a²]/(c/a)
= 3abc - b³/a²c

a and b are the roots of the equation
x² - 6x + 6 = 0
? a + b = -B/A = 6 and ab = C/A = 6
We know that,
a² + b² = (a + b)² - 2ab
= (6)² - 2 x 6 = 36 - 12 = 24
? 2(a² + b²) = 2 x 24 = 48

Since, ? and ? are the roots of the equation
8x² - 3x + 27 = 0
? ? + ? = 3/8 and ? ? = 27/8
? (?²/?)^1/3 + (?²/? )^1/8
= (?³)^1/3 + (?³)^1/3/(??)³
= ? + ? / (??)^1/3 = 3/8/(27/8)^1/3
= 3/8 x 2/3 = 1/4

When mistake is done in first degree term, the roots of the equation are -9 and -1.
? Equation
(x+ 1) (x + 9) = x² + 10x + 9 ...(i)
When mistake is done in constant term, the roots of equation are 8 and 2.
? Equation is
(x - 2) (x - 8) = x² - 10x + 16 .....(ii)
? Required equation from Eqs. (i) and (ii) is
= x² - 10x + 9
Also we see in both the cases 1st degree term is same with oposite sign i.e., in such questions we should take data from given conditions and find the correct equation.

Given that, one root of the equation
x² - bx + c = 0 is square of other root of this equation i.e., roots (?, ?²).
? Sum of roots = ? + ?² = -(-b)/1
? ? (? + 1) = b .......(i)
and product of roots= ?. ?² = c/1
? ?³ = c ? = c^1/3 ....(ii)
From Eqs. (i) and (ii).
c^1/3 (c^1/3 + 1) = b ....(iii)
On cubing both sides, we get
c(c^1/3 + 1)³ = b³
? c {c + 1 + 3c^1/3 (c^1/3 + 1)} = b³
? c {c + 1 = 3b} = b³ [from Eq. (iii)]
? b³ = 3bc + c² + c

Let ? and ? be the roots of the quadratic equation x² + px + q = 0
Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e products of roots
= q = ? . ? = 6 x 2 = 12 ...(i)
and B starts with a wrong value of q and gets the roots as 2 and - 9. But this time p is correct i.e., sum of roots
= p = ? + ? = - 9 + 2 = - 7 ..(ii)
(? - ?)² = (? + ?)² - 4? ?
= (-7)² - 4.12 = 49 - 48 = 1
[from Eqs. (i) and (ii)]
? ? - ? = 1 ..(iii)
Now, from Eqs. (ii) and (iii) , we get
? = - 3 and ? = - 4
which are correct roots .

Here, x = ??5 + 1/?5 - 1
On rationalising the terms given in square root, we get
x = ??5 + 1/?5 - 1 x ?5 + 1/?5 + 1
?5 + 1/2
Now, substituting the value of x in x² - x - 1.
? x² - x - 1 = (?5 + 1/2)² - (?5 + 1/2) - 1
= 5 + 1 + 2?5/4 - ?5 + 1/2 - 1
= 6 + 2?5 - 2?5 - 2 - 4 /4 = 0