Practice question on quadratic equation for IITJEE , Airforce and NDA Page no. 2 (Set 2)

quadratic equations

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Question no. 16

If P(x)= ax² +bx+c=0 and Q(x)= -ax² +bx+c=0, where ac≠0 then P(x).Q(x)=0 has at least

looks_one four real roots

looks_two two real roots

looks_3 four imaginary roots

looks_4 none of these

Option looks_two two real roots

ax² +bx+c=0 has D = b² -4ac , D can be less than 0 or greater than 0 , there is a possibility of having complex roots or real roots.
-ax² +bx+c=0 has D = b² + 4ac , D is always greater than or equal to 0 , so it has 2 reals roots , So P(x)Q(x) = 0 has atleast 2 real roots

Question no. 17

If the roots of the given equation 2x² +3(λ-2)x+ λ+4=0 be equal in magnitude but opposite in sign, then λ =

looks_one1

looks_two 2

looks_3 3

looks_4 2/3

option looks_two 2

let A and B are two roots of given quadratic equation
A + B = 0
3 (λ -2)/ 2 = 0
λ = 2

Question no. 18

If a+b+c=0, then the roots of the equations 4ax² +3bx+2c=0 are

looks_one equal

looks_two imaginary

looks_3 real

looks_4 none of these

option looks_3 real

D = 9b² - 4 (4a)(2c)
If a + b + c = 0 , then b= -(a+c)
b² = (a+c)²
D = 9b² - 4 (4a)(2c)
= 9(a+c)² -32ac
= 9(a² + c² + 2ac) -32ac = 9a² + 9c² - 14ac
= 9(a-c)² + 4ac > 0
D > 0

Question no. 19

If the roots of equation px²+2qx+r=0 and qx²-2√(pr) +q =0 be real, then

looks_one p=q

looks_two q²=pr

looks_3 p²=qr

looks_4 r²=pq

option looks_two q²=pr

equation 1 :
4q² - 4pr ≥ 0
q² ≥ pr (1)
Equation 2 :
4pr - 4q² ≥ 0
pr ≥ q² (2)
from 1 and 2 , only equality gives a common solution
So , q² = pr

Question no. 20

If the roots of equation ax²+x+b=0 be real, then the roots of equation x²-4√abx +1 =0 will be

looks_one rational

looks_two irrational

looks_3 real

looks_4 imaginary

option looks_4 imaginary

ax²+x+b=0 , D = 1 - 4ab ≥ 0
1 ≥ 4ab (1)
x²-4√abx +1 =0
D = 16ab - 4 = 4(4ab - 1 )
From (1)
D = 4 (4ab - 1) ≤ 0

Question no. 21

The equation x^{(3/4)(log₂x)2 +(log₂x) -5/4} =√2 has

looks_one at least one real solution

looks_two exactly three real solutions

looks_3 exactly one irrational solution

looks_4 all the above

option looks_4 all the above

Solution :
x^{(3/4)(log₂x)2 +(log₂x) -5/4} =√2
(3/4(log₂x)² +log₂x -5/4)log₂x = 1/2
log₂x = Y
(3Y²/4 + Y - 5/4 )Y = 1/2
3Y³ +4Y² - 5Y - 2 = 0
Y = 1
(Y-1)(Y² + 7Y + 2) = 0
(Y-1)(Y+2)(3Y + 1) = 0
Y = 1 , -2 , -1/3
for Y = 1 , x = 2^Y , x = 2
for Y = -2 , x = 2^-2
for Y = -1/3 , x = 2^-1/3

Question no. 22

The value of k for which 2x² -kx +x+8=0 has equal and real roots are

looks_one-9, -7

looks_two 9,7

looks_3 -9,7

looks_4 9,-7

option looks_4 9,-7

Solution :
b² - 4ac = 0
(-k + 1)² -4(8)(2) = 0
k² + 1 -2k -64 = 0
k² -2k -63 =0
(k-9)(k+ 7) = 0
k = 9 , -7

Question no. 23

If the roots of equation x²-8x+(b² -6b)=0 be real ,then

looks_one -2< b <8

looks_two 2 < b <8

looks_3 -2≤b≤8

looks_4 2≤b≤8

option looks_3 -2≤b≤8

Solution :
x²-8x+(b² -6b)=0
D ≥ 0
64 - 4(b² -6b) ≥ 0
16 - b² +6b ≥ 0
b² -6b - 16 ≤ 0
b² -8b +2b - 16 ≤ 0
(b-8)(b+2)≤0
b= [-2 , 8]

Question no. 24

Roots of the equation 2x²-5x+1=0, x²+5x+2=0 are

looks_one reciprocal and of same sign

looks_two reciprocal and of opposite sign

looks_3 equal in product

looks_4 none of these

option looks_two reciprocal and of opposite sign

Solution:
2x²-5x+1=0
x = 5/4 +√17/4 , 5/4 -√17/4
x²+5x+2=0
x = -5/2 -√17/2 , -5/2 +√17/2
5/4 +√17/4 is reciprocal and opposite of -5/2 -√17/2

Question no. 25

The expression x²+2bx+c has the positive value if

looks_one b²-4c>0

looks_two b²-4c b <0

looks_3 c²< (b)

looks_4 b²<(c)

option looks_4 b²<(c)

Solution
x²+2bx+c
coefficient of x² is +ve
for Positive value : b² - 4ac < 0
4b² - 4c < 0
b² < c

Question no. 26

what is the value of ?

looks_one 5

looks_two √5

looks_3 1

looks_4 5^1/4

option looks_one 5

solution :

x = √5x
x² = 5x
x = 0 , 5

Question no. 27

If roots of the equation 4x²+px+9=0 are equal, then p=

looks_one 144

looks_two 12

looks_3 -12

looks_4 ∓12

option looks_4 ∓12

p² - 4.4.9 = 0
p² = 4.4.9
p = 12 , -12

Question no. 28

if x= 2+2^1/3 +2^2/3 , then what is the value of x³ -6x²+6x ?

looks_one 1

looks_two 2

looks_3 3

looks_4 -2

option looks_two 2

x = 2+2^1/3 +2^2/3
(x-2)³ = ( 2^1/3 +2^2/3 ) ³
x³ - 8 + 12x - 6x² = 2 + 2² + 3.2^1/3.2^2/3(2^1/3 +2^2/3)
x³ + 12x - 6x² = 8 + 6 + 3.2 (x-2)
x³ + 12x - 6x² = 14 -12 + 6x
x³ + 6x - 6x² = 2

Question no. 29

The roots of equation (x-p)(x-q)=r² where p,q,r are real ,are

looks_one always complex

looks_two always real

looks_3 always purely imaginary

looks_4 none of these

option looks_two always real

(x-p)(x-q) = r²
x² -(p+q)x + pq = r²
x² - (p+q)x +pq -r² = 0
D = ( p + q)² - 4(pq -r²)
D = p² + q² + 2pq - 4pq + 4r²
D = (p - q)² + 4r² ≥ 0

Question no. 30

The equations has

looks_one no roots

looks_two one root

looks_3 two equal roots

looks_4 infinite roots

option looks_one no roots

If we solve the equation using algebra , we find x = 1 , but x = 1 can't be a solution , So no roots

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