Complex Numbers and Quadratic Equation Test - 4

For Real roots
D = b²  -4ac ≥0
or b²  ≥4ac
For eqn(1); k²  ≥4.1.64 ⇒k ≥16 ……….(I)
For eqn (2) (-8)²  ≥4.1.k ⇒k ≤16 ……….(II)
Clearly k = 16 satisfies (I) &(II)
∴(b) is correct.

If one root = 2 + √5
Its other root = 2 –√5 (Irrational conjugate) Eqn is
x²  –(Sum of roots) x + Product of roots = 0
or x²  –(2 + √5 + 2 –√5)x + (2 + √5)(2 –√5)=0
or x²  –4x + (4 –5) = 0
or x²  –4x –1 = 0
(b) is correct.

Let 1st root = 1
Then 2nd root = 2
The Eqn. is x²  –(1 + 2) + 1 ×2 = 0
or x²  –3x + 2 = 0
Comparing it with ax²  + bx + c = 0 we get;
a = 1; b = -3 ; c = 2
Go by choices (GBC)
(a) 2b²  = 3ac
2.(-3)²  =3.1.2 (False)
(c) 2b²  =9.ac
2. (-3)²  = 9.1.2
⇒18 = 18 (True)
Hence, Option (c) is (true)

Quadratic Eqn. having roots αand βis
x²  –(α+ β)x + αβ= 0
or ; x²  –(- 2)x + (- 3)= 0
or; x²  + 2x –3 = 0
(b) is correct.

∵Roots are in A.R
Let roots are a –d; a; a + d
So, (a –d)+a + (a + d) = 15
or; 3a = 15
or; a = 5
And Product of roots
(a –d ). a . (a + d ) = 45
or (5 –d);5. (5 + d) = 45
or 25 –d²  = 9
or; d²  = 25 –9 = 16
or; d = √16 = 4
Hence; roots are
a –d, a, a + d = 5 –4; 5; 5 + 4
= 1; 5 ; 9.
The value of K
= Sum of product of two roots in a order
= (1 ×5) + (5 ×9) + (9 ×1)
= 5 + 45 + 9 = 59
(b) is correct.

∵x²  + 3x –4 = 0
or; x²  –4x + x –4 = 0
or; x(x –4) + 1(x –4) = 0
or; (x –4)(x + 1) = 0
x = 4; -1
Eqn. having roots  1/2  & 1/−1  = 1/4  &–1 is.
or x²  –(1/4  –1) + 1/4(-1) = 0
or x²  + 3/4x – 1/4  = 0
Multiplying by 4 ; we get
4x²  + 3x -1 = 0
Comparing it with kx²  + 3x -1 = 0
We get K = 4
Tricks : Eqn. having roots the reciprocal of the roots of ax²  + bx + c = 0 is cx²  + bx +a = 0 i.e. 1st and last term interchanges.

let α, βare roots of x²  –kx + 8 = 0
∴α+ β= -b/a = −(−k)/1  = k &α. β= c/a = 8/1 = 8
(α–β)²  = (α+ β)²  –4 αβ= 4²
⇒k²  –4 ×8 = 16
or k²  = 48 ⇒k = ±√16 ×3  ⇒k = ±4 √3
(d) is correct.

(b) is correct
Quadratic Eqn. is
x²  –(α+ β)x + αβ= 0
x²  –(-2)x + (-3) = 0
∴x²  + 2x –3 = 0

y³  + y²  –y –1 = 0
or y² (y + 1) –1(y + 1) = 0
or (y + 1)(y²  –1) = 0
or(y + 1)(y + 1)(y –1)=0
y = -1 ; -1; 1.

Roots are 2 + √3 and 2 –√3 (conjugate of 2 + √3)
Eqn is
x²  –(Sum of roots) x + product of roots = 0
x²  –4x + (4 –3) = 0
x²  + px + a = 0
∴P = -4 ; a = 1