Two students A and B solve an equation of the from x ² + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. B starts with a wrong value of q and gets the roots as 2 and -9. What are the correct roots of the 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

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.

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, 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

According to question , we have
Given equation is :- 2^2x - 3 × 2^x × 2² + 32 = 0
? 2^2x - 12 × 2^x + 32 = 0
? y² - 12y + 32 = 0, where 2^x = y
? ( y - 8 ) ( y - 4 ) = 0
? y = 8, y = 4
? 2^x = 8 or, 2^x = 4
? 2^x = 2³ or, 2^x = 2²
? x = 3 or, x = 2.
Therefore , the roots of the equation are 3 and 2.