Test 4 Ch. 4 Quadratic Equations

Test 4

Ch. 4

Quadratic Equations

Q. Each root of x²– bx + c = 0 is decreased by 2. The resulting equation is x²+2x + 1 = 0, then

(a) b = 6, c = 9    (b) b = 3, c = 5

(c) b = 2, c = –1 (d) b –4, c = 3

Sol.

For x²– bx + c = 0 we have

α + β = b/1=b

2 + 2 =b

b = 4

α×β = c/1=c

2×2 =c

c = 4

For x²+2x + 1 = 0 we have

α' + β' = –2/1=–2

α'×β' = 1/1=1

Now ATQ

α + β = α'+ β' 

b= −2

again 

α × β = α' × β' 

 c = 1

For x x2 1 0 2 − + = we have

2 = b − 4 & b = 6

and 1 = c b − + 2 4

 = c − + 2 6 # 4

 = c − 8

c = 1 8 + = 9

Thus (a) is correct option.

16. The condition for one root of the quadratic equation ax²+ bx + c = 0 to be twice the other, is

(a) b² = 4ac

(b) 2b²=9ac

(c) c²= 4a + b²

(d) c²= 9a – b²

Sol.

ax²+ bx + c = 0

α = α

β = 2α

Sum of zeroes

α + β = –b/a 

α +2α =−b/a

3α =−b/a

α =−b/3a

Product of zeroes

α × β = c/a 

α × 2α =c/a

2 α² = c/a 

2(–b/3a)²= c/a

 2b

c =

2 9 ab a c 2 2 - = 0

a b2 9ac 2 ^ h - = 0

Since, a ! 0, 2b2 = 9ac

Hence, the required condition is 2 9 b ac 2 = .

Thus (b) is correct option.