The roots of the equation 4x - 3 × 2^x × 2² + 32 = 0 would include :

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

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 .

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

As per the given above question , we have
From equation ?. x² - 24x + 144 = 0
? x² ? 12x - 12x + 144 = 0
? x(x ? 12) ? 12(x ? 12) = 0
? (x ? 12)² = 0
? x = 12
From equation ?. y² ? 26y + 169 = 0
? y² ? 13y ? 13y + 169 = 0
? y(y ? 13) ? 13(y ? 13) = 0
? (y ? 13)² = 0
? y = 13
Hence, required answer will be x < y .