Page 4 - CHAPTER 4 (Quadratic equations)

CHAPTER 4

QUADRATIC EQUATIONS

2 2 5
⇒ -1+2 and - +2
25 1 1
4 2
2 27
⇒ + 1 and 8-10+2 ⇒ and 0
25 25

1
i.e., x=5 does not satisfy equation but = satisfies.
2
5
2
1
So, x=5 is not a solution but = is a solution of - +2=0.
2 x 2 x
b
a
ii. Putting x= and x= in the given equation.
b a
a a b b
a² ( ) ²-3ab ( ) +2b ² and a² ( ) ²-3ab ( ) +2b²
b b a a
a
⇒ +2b²-3a² and 0
b 2

a
b
i.e., x= does not satisfy but x= satisfies the given equation.
b a
b
a
Hence x= is a solution, but x= is not a solution of a²x²-3abx+2b²=0.
a b
3
Example 3: Find the values of l and m for which x= and x=-2 are the
4
roots of the equation lx²+mx-6=0.

3
Solution: Since x= and x=-2 are the roots of the equation lx²+mx-6=0.
4

2
3
3
2
∴ (l ( ) +m ( ) -6) =0 and (l(-2) + m(-2)- 6)=0
4 4
3
⇒l× 9 + m × - 6=0 and 4l – 2m – 6 = 0
16 4

⇒ 9l+12m-96 =0 and 4l – 2m – 6 = 0
16

⇒ 9l + 12m – 96 = 0 and 4l – 2m – 6 = 0

⇒ 3l + 4m – 32 = 0 …. (i)

and 2l – m – 3 = 0 …. (ii)

Multiplying (2) by 4, we get 8l –4m – 12 = 0 …. (iii)

Adding (1) and (3), we get l = 4

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