One root of the quadratic equation x² - 5 x + 6 = 0 is 3. Find the other root.

Given quadric equation is
x² + x/b + 1/c = 0 ...(i)
Now, by condition the roots of the Eq.(i) are ? and 1/?.
Now,product of roots = (1/c) / (1/a)
? ?.(1/?) = a/c
? c = a
which is the required relation.

2x² - 7xy + 3y² = 0
? 2x² - 6xy - xy + 3y² = 0
? 2x(x - 3y) - y(x - 3y) = 0
? (2x - y) (x - 3y) = 0
Either, 2x - y = 0 ? 2x = y
? x/y = 1/2
or x - 3y = 0
? x = 3y ? x/y = 3/1

[a + (1/a)]² = 3
Taking square roots both sides, we get
a + (1/a) = ?3
On cubing both sides, we
[a + (1/a)]³ = (?3)³
? a³ + 1/a³ + 3.a.1/[a(a + 1/a)] = 3 ?3
? a³ + 1/a³ + 3?3 = 3?3
? a³ + 1/a³ = 0

2x² -11x + 15 = 0
[by factorisation method]
? 2x² - (6x + 5x) + 15 = 0
? 2x² - 6x - 5x + 15 = 0
? 2x(x - 3) - 5 (x - 3) = 0
? (2x - 5) (x - 3) = 0
? x = 5/2, 3
Hence, the roots are 5/2 and 3.

Given, one root of x² - 6kx + 5 = 0 is 5.
? x = 5 satisfies x² - 6kx + 5 = 0
? 5² - 30k + 5 = 0
? 25 - 30k + 5 = 0
? 30 - 30k = 0
? 30k = 30
? k = 1

Given that :- sum of the roots = 6
And Product of two roots = -16
The required quadratic equation is x² ? (sum of the roots) x + (product of the roots) = 0
? x² - 6x - 16 = 0

According to question ,we can say that
?. 4x + 7y = 209 ...............( 1 )
?. 12x ? 14y = ? 38 .................... ( 2 )
Multiplying (1) by (2):
8x + 14y = 418 ................(3)
Adding (2) and (3):
20x = 380 ? x = 19
Substituting the value of x in (1), we get
76 + 7y = 209
? 7y = 133 ? y = 19
From above equations it is clear that x = y is correct answer .