Quadratic Equation Questions for NMAT – Download [PDF]

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NMAT Quadratic Equation Questions
NMAT Quadratic Equation Questions

Quadratic Equation Questions for NMAT PDF:

Download Quadratic Equation Questions for NMAT PDF. Top 10 very important Quadratic Equation Questions for NMAT based on asked questions in previous exam papers.

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Question 1: The number of integers that satisfy the equality (x25x+7)x+1=1 is

a) 3

b) 2

c) 4

d) 5

Question 2: The number of distinct real roots of the equation (x+1x)23(x+1x)+2=0 equals

Question 3: How many disticnt positive integer-valued solutions exist to the equation (X27x+11)(X213x+42)=1 ?

a) 8

b) 4

c) 2

d) 6

Question 4: If x2+x+1=0, then x2018+x2019 equals which of the following:

a) x+1

b) x

c) x

d) x1

Question 5: If U2+(U2V1)2= −4V(U+V) , then what is the value of U+3V ?

a) 0

b) 12

c) 14

d) 14

Question 6: If x+1=x2 and x>0, then 2x4  is

a) 6+45

b) 3+35

c) 5+35

d) 7+35

Question 7: If x2+3x-10is a factor of 3x4+2x3ax2+bxa+b4 then the closest approximate values of a and b are

a) 25, 43

b) 52, 43

c) 52, 67

d) None of the above

Question 8: If xy+yz+zx=0, then (x+y+z)2 equals

a) (x+y)2+xz

b) (x+z)2+xy

c) x2+y2+z2

d) 2(xy+yz+xz)

Question 9: If the equation x3ax2+bxa=0 has three real roots, then it must be the case that,

a) b=1

b) b 1

c) a=1

d) a 1

Question 10: If the roots of the equation x3ax2+bxc=0 are three consecutive integers, then what is the smallest possible value of b?[CAT 2008]

a) 13

b) 1

c) 0

d) 1

e) 13

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Answers & Solutions:

1) Answer (A)

(x25x+7)x+1=1

There can be a solution when (x25x+7)=1 or x25x +6=0

or x=3 and x=2

There can also be a solution when x+1 = 0 or x=-1

Hence three possible solutions exist.

2) Answer: 1

Let a=x+1x
So, the given equation is a23a+2=0
So, a can be either 2 or 1.

If a=1, x+1x=1 and it has no real roots.
If a=2, x+1x=2 and it has exactly one real root which is x=1

So, the total number of distinct real roots of the given equation is 1

3) Answer (D)

(X27x+11)(X213x+42)=1

if (X213x+42)=0 or (X27x+11)=1 or (X27x+11)=-1 and (X213x+42) is even number

For X=6,7 the value (X213x+42)=0

(X27x+11)=1 for X=5,2.

(X27x+11)=-1 for X=3,4 and for X=3 or 4, (X213x+42) is even number.

.’. {2,3,4,5,6,7} is the solution set of X.

.’. X can take six values.

4) Answer (C)

We know that,

x31=(x1)(x2+x+1)

Since, x2+x+1=0

 x31 = 0

=> x3 = 1

Now, x2018+x2019

(x3)672x2(x3)673

1672x21673

x2 + 1

= -x

Hence, option C.

5) Answer (C)

Given that U2+(U2V1)2= −4V(U+V)

U2+(U2V1)(U2V1)= −4V(U+V)

U2+(U22UVU2UV+4V2+2VU+2V+1) = −4V(U+V)

U2+(U24UV2U+4V2+4V+1) = −4V(U+V)

2U24UV2U+4V2+4V+1=4UV4V2

2U22U+8V2+4V+1=0

2[U2U+14]+8[V2+V2+116]=0

2(U12)2+8(V+14)2=0

Sum of two square terms is zero i.e. individual square term is equal to zero.

U12 = 0 and V+14 = 0

U = 12 and V = 14

Therefore, U+3V12+13414. Hence, option C is the correct answer.

6) Answer (D)

We know that x2x1=0
Therefore x4=(x+1)2=x2+2x+1=x+1+2x+1=3x+2
Therefore, 2x4=6x+4

We know that x>0 therefore, we can calculate the value of x to be 1+52
Hence, 2x4=6x+4=3+35+4=35+7

7) Answer (C)

If x2+3x-10is a factor of 3x4+2x3ax2+bxa+b4
Then x = -5 and x = 2 will give 3x4+2x3ax2+bxa+b4 = 0
Substituting x = -5 we get,
3(5)4+2(5)3a(5)2+b(5)a+b4=0
Solving we get,
26a+4b=1621…….(i)
Substituting x = 2 we get,
3(2)4+2(2)3a(2)2+b(2)a+b4=0
=> 5a3b=60……..(ii)
Solving i and ii we get
a and b 52,67
Hence, option C is the correct answer.

8) Answer (C)

(x+y+z)2=x2+y2+z2+2(xy+yz+xz)
as xy+yz+xz=0
so equation will be resolved to x2+y2+z2

9) Answer (B)

It can be clearly seen that if b=1 then x2(xa)+(xa)=0 an the equation gives only 1 real value of x

10) Answer (B)

b = sum of the roots taken 2 at a time.
Let the roots be n-1, n and n+1.
Therefore, b=(n1)n+n(n+1)+(n+1)(n1)=n2n+n2+n+n21
b=3n21. The smallest value is -1.

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