Page 4 - LN

Hence, the solution of the given quadratic equation are and .

2 2 2
2
2
(iii) a b x - (a + b )x + 1 = 0 , a 0, b 0
Ans: The given quadratic equation is
2
2
2 2 2
a b x - (a + b )x + 1 = 0, a 0, b 0
2
2
2 2 2
a b x - a x - b x + 1 = 0
2
2
2
a x(b x - 1)- 1(b x - 1) = 0
2
2
(a x - 1)(b x - 1) = 0
2
2
a x - 1 = 0 or b x - 1 = 0
x = or x =
Thus, the two solution of the given quadratic equation are and

Example 4: Sides (in cms) of a right triangle are x - 1, x and x + 1. Find the sides of the
triangle.
Ans: Three sides of a right triangle are (x - 1), x and (x + 1).
The longest side (hypotenuse) is (x + 1).
Using Pythagoras Theorem, we have,

x = 0 or x = 4
But x 0 (since length of a side cannot be zero)
x = 4
Now, x - 1 = 4 - 1 = 3
and x + 1 = 4 + 1 = 5
Hence, the three sides of triangle are 3 cm, 4 cm, and 5 cm.

4