Therefore, 22 0 2 0
4 2
±
x = =± i.e., x =
1
2
So, the roots are 1 1
, .
2 2
Example-12.
Find the roots of the following equations:
(i)
1
x x += ¹ 3, 0
x
(ii)
1 1 3, 0, 2
2
- =¹
-
x
x x
Solution :
(i) 1
x + =3
x
. Multiplying both sides of equation by x, we get
x²+1=3x
i.e., x
2
– 3x + 1 = 0, which is a quadratic equation.
Here, a = 1, b = – 3, c = 1
So, b
2
– 4ac = 9 – 4 = 5 > 0
Therefore, 3 5
2
±
x = (why ?)
So, the roots are 3 5
2
+
and 3 5
.
(ii)
1 1 3, 0, 2.
As x ¹ 0, 2, multiplying the equation by x (x – 2), we get
(x – 2) – x = 3x (x – 2)
= 3x
2
– 6x
So, the given equation reduces to 3x
2
– 6x + 2 = 0, which is a quadratic equation.
Here, a = 3, b = – 6, c = 2. So, b
2
– 4ac = 36 – 24 = 12 > 0
Therefore, 6 12 6 2 3 3 3
pgno.121