In solving a problem, one student makes a mistake in the coefficient of the first degree term and obtain -9 and -1 for the roots. Another student makes a mistake in the constant term of the equation and obtains 8 and 2 for the roots. The correct equation was

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

Let ? and ? be the roots of the equation
x² + px + q = 0
Then, ? + ? = -p, ? ? = q
According to the question,
? + ? = ?² + ?²
? ? + ? = (? + ?)² - 2? ?
? -p = p² - 2q ? p² + p = 2q

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 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