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
The given equation is
x2 - 5x + 6 = 0 ? x 2 - 2x - 3x+ 6 = 0
? x( x - 2 ) - 3( x - 2 ) = 0
? ( x - 2 ) ( x - 3 ) = 0
? x - 2 = 0 or x - 3 = 0
? x = 2 or x = 3
Thus, the other root of the given quadratic equation is 2.
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.
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 .
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