Practice question on quadratic equations for IIT-JEE , NDA and Airforce (Set 1)

quadratic equations

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Question no. 1

Let one root of ax² +bx+c=0 , where a ,b ,c are integers, be 3+ √5 ,then the other root is

looks_one 3-√5

looks_two 3

looks_3 √5

looks_4 none of these

Option looks_one 3-√5

Whenever roots are in complex or in square roots form , then they are conjugate to each other
therefore , the other root is 3-√5

Question no. 2

The roots of the given equation (p-q)x² +(q-r)x+(r-p)=0

looks_one (p-q)/(r-p),1

looks_two (q-r)/(p-q),1

looks_3 (r-p)/(p-q),1

looks_4 (q-r)/(q-p),1

looks_3 (r-p)/(p-q),1

(p-q)x² +(q-r)x+(r-p)=0

Question no. 3

If one root of equations x² +px+12=0 is 4 ,while the root of equations x² +px+q=0 are equal, then the value of q is

looks_one 4

looks_two 4/49

looks_3 49/4

looks_4 none of these

Option looks_3 49/4

x² +px+12=0
(4)² + 4p + 12 = 0
p = -7
Roots are equal , therefore D =0
b² - 4ac = 0
p² - 4 *1 *q = 0
49 - 4q = 0
q = 49/4

Question no. 4

The solution of equation will be

looks_one 2,-1

looks_two 0,-1,-1/5

looks_3 -1,-1/5

looks_4 none of these

Option looks_4 none of these

x + 1/x = 2
x² - 2x + 1 =0
(x - 1)² = 0
x = 1 , 1

Question no. 5

If, then x is

looks_one 2

looks_two 3

looks_3 6

looks_4 5

Option looks_3 6

Solution :

Question no. 6

The roots of the equation 2^x+227^x/(x+1)=9 are

looks_one 1-log₂3, 2

looks_two 1, log₂(2/3)

looks_3 2, -2

looks_4 -2, 1-log3/log2

Option looks_4 -2, 1-log3/log2

Solution :

Question no. 7

Let α and β be the roots of equation x² +x+1=0 ,the equation whose roots are α¹⁹ , β⁷

looks_one x² -x-1=0

looks_two x² -x+1=0

looks_3 x² +x-1=0

looks_4 x² +x+1=0

Option looks_4 x² +x+1=0

Solution :
α =ω , β = ω²
α¹⁹ = ω¹⁹ = ω(ω³)⁶ = ω
β⁷ = (ω³)⁴ω² = ω²
So ω and ω² are the roots of x² +x+1=0

Question no. 8

The number of solutions is

looks_one 2

looks_two 3

looks_3 1

looks_4 none of these

Option looks_4 none of these

Solution :

log5 = 2log(x-2) - log(x² -1)
log5 = log (x-2)²/(x² -1)
5 = (x-2)²/(x² -1)
5x² - 1 = x² + 4 - 4x
4x² + 4x -5 = 0
x = 0.724 , x = 1.72 But x must be greater than 2

Question no. 9

If log₂x +log_x2=10/3 =log₂y +log_y2 , x≠y , then x+y=

looks_one 2

looks_two 65/8

looks_3 37/6

looks_4 none of these

Option looks_4 none of these

Solution :
log₂x +log_x2=10/3
log₂x +1/log_x2=10/3
(log₂x)² +1 = 10(log_x2)/3
let Z = log_x2
3Z^2 -10Z + 3 = 0
Z = 3 , 1/3
log₂x = 3 ⇒ x = 8
log₂x = 1/3 ⇒ x = 2^1/3 x = 8 , y = 2^1/3
x + y = 8 + 2^1/3

Question no. 10

The value of is

looks_one -1

looks_two 1

looks_3 2

looks_4 3

Option looks_3 2

Solution :

x = √(2 + x)
x² = 2 + x
x² - x -2 = 0
(x-2)(x-1) = 0
x = 2 , -1
-1 can't be solution

Question no. 11

The equation √(x+1)- √(x-1)=√(4x-1) has

looks_one no solution

looks_two one solution

looks_3 two solution

looks_4 more than two solution

option looks_one no solution

Solution :
√(x+1)- √(x-1)=√(4x-1)
Squaring both sides
x+1 + x-1 -2√(x²-1) = 4x -1
-2√(x²-1) = 2x -1
Squaring both sides
4(x²-1) = 4x² + 1 -4x
-5 = -4x
x = 5/4 but this value does not satisfy the equation

Question no. 12

A real root of the equation log₄{ log₂(√x+8-√x)}=0 is

looks_one 1

looks_two 2

looks_3 3

looks_4 4

option looks_one 1

log₄{ log₂(√x+8-√x)}=0
log₂(√x+8-√x) = 1
√x+8-√x = 2
√x+8 = 2 + √x
squaring both sides
x + 8 = 4 + x + 4√x
4 = 4√x
x = 1

Question no. 13

The roots of | x-2|² + |x-2| -6=0 are

looks_one 0,4

looks_two -1, 3

looks_3 4,2

looks_4 5,1

Option looks_one 0,4

| x-2|² + |x-2| -6=0 Case 1 : x-2 > 0
(x-2)² + (x-2) -6=0
x² -4x + 4 + x-2 -6 =0
x² -3x - 4 =0
x² -4x + x - 4 = 0
(x+1)(x-4)= 0
x= -1 , 4 [only 4 is solution]
Case 2 :
x-2 < 0
(x-2)² - (x-2) - 6 =0
x² + 4 - 4x -x + 2 -6 = 0
x² -5x = 0
x = 0 , 5 [only 0 is solution]

Question no. 14

The numbers of solution for the equation x² -5|x| +6=0 is

looks_one 4

looks_two 3

looks_3 2

looks_4 1

Option looks_one 4

x² -5|x| +6=0
There will be two cases for x > 0 , x < 0
Case 1 : x > 0
x² -5x + 6 =0
x² - 2x - 3x + 6 =0
(x-2)(x-3)=0
x =2 , x= 3 [both solution, x must be greater than zero]
Case 2 : x < 0
x² +5x + 6 =0
x² +2x +3x +6 =0
(x+2)(x+3)= 0
x= -2 , x= -3 [Both solution , x must be less than zero]

Question no. 15

The numbers of solution for the equation x² +5|x| +6=0 is

looks_one 4

looks_two 2

looks_3 3

looks_4 0

Option looks_4 0

x² +5|x| +6=0
There will be two cases for x > 0 , x < 0
Case 1 : x > 0
x² +5x + 6 =0
x² + 2x + 3x + 6 =0
(x+2)(x+3)=0
x =-2 , x= -3 [No solution, x must be greater than zero]
Case 2 : x < 0
x² -5x + 6 =0
x² -2x -3x +6 =0
(x-2)(x-3)= 0
x= 2 , x= 3 [No solution , x must be less than zero]

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