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Chapter 2 Theory of Quadratic Equation


Updated: 09 Apr 2023

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In Chapter 2, maths class 10, students will learn about the theory of quadratic equations (Discriminant of quadratic equations & Cube Roots of Unity). This chapter also includes multiple-choice questions (MCQs of Chapter 2 Mathematics) to learn and test their understanding of the material to secure good marks.

Chapter 2 Maths Class 10 MCQs

Exercise # 2.1 MCQs

Discriminant of a Quadratic Equation

In this portion of Chapter 2 Maths Class 10 includes the MCQs of Discriminant of a Quadratic Equation.

1. In quadratic formula, the expression b^{2}-4 a c is called ________ of quadratic equation.
(a) Factorization
(b) Functions
(c) Discriminant
(d) Irrational

Answer:
Discriminant


2. The value of the discriminant is used to determine the number of solutions of a ________ equation.
(a) Linear
(b) Quadratic
(c) Simultaneous
(d) None of these

Answer:
Quadratic


3. If b^{2}-4 a c ________ then the roots are real, equal and rational.
(a) =0
(b) <0
(c) >0
(d) None of these

Answer:
=0


4. If b^{2}-4 a c ________ then the roots are unequal and imaginary.
(a) =0
(b) <0
(c) >0
(d) None of these

Answer:
<0


5. If b^{2}-4 a c ________ and root is a perfect square, then roots are real, unequal and rational.
(a) =0
(b) <0
(c) >0
(d) None of these

Answer: >0


6. If b^{2}-4 a c ________ and root is not a perfect square, then roots are real, unequal and irrational.
(a) =0
(b) <0
(c) >0
(d) None of these

Answer: >0


7. If b^{2}-4 a c>0 and roots are perfect square then roots are ________
(a) Rational
(b) Irrational
(c) Equal
(d) Imaginary

Answer:
Rational


8. If b^{2}-4 a c>0 and roots are not perfect square then roots are ________
(a) Rational
(b) Irrational
(c) Equal
(d) Imaginary

Answer:
Irrational


9. The discriminant of x^{2}+9 x+2=0
(a) -73
(b) 73
(c) 0
(d) 9 x+2

Answer: 73
Explanation:
x^{2}+9 x+2=0
Compare it with a x^{2}+b x+c=0
Here a=1, b=9, c=2
As we have
Discriminant =b^{2}-4 a c
Discriminant =(9)^{2}-4(1)(2)
Discriminant =81-8
Discriminant =73

Nature of Roots

10. What is the nature of the roots of x^{2}-8 x+16=0
(a) Real, equal and rational
(b) Real, unequal and irrational
(c) Real, unequal and rational
(d) Unequal and imaginary

Answer: Real, equal and rational
Explanation:
x^{2}-8 x+16=0
Compare it with a x^{2}+b x+c=0
Here a=1, b=-8, c=16
As we have
Discriminant =b^{2}-4 a c
Discriminant =(-8)^{2}-4(1)(16)
Discriminant =64-64
Discriminant =0
Thus the roots are real, equal and rational


11. What is the nature of the roots of x^{2}+9 x+2=0
(a) Real, equal and rational

(b) Real, unequal and irrational
(c) Real, unequal and rational
(d) Unequal and imaginary

Answer: Real, unequal and irrational
Explanation:
x^{2}+9 x+2=0
Compare it with a x^{2}+b x+c=0
Here a=1, b=9, c=2
As we have
Discriminant =b^{2}-4 a c
Discriminant =(9)^{2}-4(1)(2)
Discriminant =81-8
Discriminant =73>0
Thus the roots are real, unequal and irrational


12. What is the nature of the roots of 6 x^{2}-x-15=0
(a) Real, equal and rational
(b) Real, unequal and irrational
(c) Real, unequal and rational
(d) Unequal and imaginary

Answer: Real, unequal and rational
Explanation:
6 x^{2}-x-15=0
Compare it with a x^{2}+b x+c=0
Here a=6, b=-1, c=-15
As we have
Discriminant =b^{2}-4 a c
Discriminant =(-1)^{2}-4(6)(-15)
Discriminant =1+360
Discriminant =361
Discriminant =19^{2}>0
Thus the roots are real, unequal and rational


13. What is the nature of the roots of 4 x^{2}+x+1=0
(a) Real, equal and rational
(b) Real, unequal and irrational
(c) Real, unequal and rational
(d) Unequal and imaginary

Answer: Unequal and imaginary
Explanation:
4 x^{2}+x+1=0
Compare it with a x^{2}+b x+c=0
Here a=4, b=1, c=1
As we have
Discriminant =b^{2}-4 a c
Discriminant =(1)^{2}-4(4)(1)
Discriminant =1-16
Discriminant =-15 <0
Thus the roots are unequal and imaginary


14. Without solving, determine the nature of the roots of the 3 x^{2}-4 x+6=0
(a) Real, equal and rational
(b) Real, unequal and irrational
(c) Real, unequal and rational
(d) Unequal and imaginary

Answer: Unequal and imaginary
Explanation:
3 x^{2}-4 x+6=0
Compare it with a x^{2}+b x+c=0
Here a=3, b=-4, c=6
As we have
Discriminant =b^{2}-4 a c
Discriminant =(-4)^{2}-4(3)(6)
Discriminant =16-72
Discriminant =-56 <0
Thus the roots are unequal and imaginary


15. Determine the value of {k} for which the given quadratic equation have real roots.
k x^{2}+4 x+1=0

(a) k>4
(b) k<4
(c) k \geq 4
(d) k \leq 4

Answer:
Explanation:
k x^{2}+4 x+1=0
Compare it with a x^{2}+b x+c=0
Here a=k, b=4, c=1
If roots are Real
Discriminant =b^{2}-4 a c \geq 0
b^{2}-4 a c \geq 0
(4)^{2}-4(k)(1) \geq 0
16-4 k \geq 0
16 \geq 4 k
\frac{16}{4} \geq k
4 \geq k
k \leq 4


16. For what value of {k} the roots of the following equation are imaginary 2 x^{2}+3 x+k=0
(a) k>4
(b) k<4
(c) k \leq \frac{9}{8}
(d) k>\frac{9}{8}

Answer: k> \frac{9}{8}
Explanation:
2 x^{2}+3 x+k=0
Compare it with a x^{2}+b x+c=0
Here a=2, b=3, c=k
If roots are Imaginary
Discriminant =b^{2}-4 a c <0
b^{2}-4 a c <0
(3)^{2}-4(2)(k) <0
9-8 k <0
9<8k
\frac{9}{8}
k> \frac{9}{8}


17. Quadratic equation ax^2+bx+c=0 has equal roots if b^2-4ac=
(a) =0
(b) <0
(c) >0
(d) None of these

Answer:
=0

Exercise # 2.2 MCQs

In this exercise of Chapter 2 Maths Class 10 includes the MCQs of Cube Root of Unity.

Cube Root of Unity

1. The cube root of unity are
(a) 1
(b) \frac{-1+i \sqrt{3}}{2}
(c) \frac{-1-i \sqrt{3}}{2}
(d) All of these

Answer:
All of these


2. The cube root of unity are
(a) 1
(b) \omega
(c) \omega^{2}
(d) All of these

Answer:
All of these


3. The sum of cube roots of unity is ________
(a) Zero
(b) 1
(c) \alpha+\beta
(d) \alpha \beta

Answer: Zero
Explanation:
As \omega=\frac{-1+i \sqrt{3}}{2}
And \omega^{2} \frac{-1-i \sqrt{3}}{2}
1+\omega+\omega^{2}
=1+\frac{-1+i \sqrt{3}}{2}+\frac{-1-i \sqrt{3}}{2}
=\frac{2+(-1+i \sqrt{3})+(-1-i \sqrt{3})}{2}
=\frac{2-1+i \sqrt{3}-1-i \sqrt{3}}{2}
=\frac{2-1-1+i \sqrt{3}-i \sqrt{3}}{2}
=\frac{1-1}{2}
=\frac{0}{2}
=0


4. 1+\omega+\omega^{2}=
(a) Zero
(b) 1
(c) \alpha+\beta
(d) \alpha \beta

Answer: Zero
Explanation:
See MCQS No. 3


5. 1+\omega=
(a) \omega
(b) \omega^{2}
(c) -\omega
(d) -\omega^{2}

Answer: -\omega^{2}
Explanation:
1+\omega+\omega^{2}=0
1+\omega=\omega^{2}


6. 1+\omega^{2}=
(a) \omega
(b) \omega^{2}
(c) -\omega
(d) -\omega^{2}

Answer: -\omega
Explanation:
1+\omega+\omega^{2}=0
Then
1+\omega^{2}=-\omega


7. \omega+\omega^{2}=
(a) \omega
(b) \omega^{2}
(c) 1
(d) -1

Answer: -1
Explanation:
1+\omega+\omega^{2}=0
Then
\omega+\omega^{2}=-1


8. The Product of cube roots of unity is ________
(a) Zero
(b) 1
(c) \alpha+\beta
(d) \quad \alpha \beta

Answer: 1
Explanation:
As \omega=\frac{-1+i \sqrt{3}}{2}
And \omega^{2} \frac{-1-i \sqrt{3}}{2}
1.\omega \cdot \omega^{2}
=1 \cdot\left(\frac{-1+i \sqrt{3}}{2}\right)\left(\frac{-1-i \sqrt{3}}{2}\right)
=\frac{(-1+i \sqrt{3})(-1-i \sqrt{3})}{2 \times 2}
=\frac{(-1)^{2}-(i \sqrt{3})^{2}}{4}
=\frac{1-i^{2}(3)}{4}
=\frac{1-3 i^{2}}{4}
=\frac{1-3(-1)}{4} \qquad i^{2}=1
=\frac{1+3}{4}
=\frac{4}{4}
=1


9. 1 . \omega \cdot \omega^{2}=
(a) Zero
(b) 1
(c) \alpha+\beta
(d) \alpha \beta

Answer:
1


10. \omega^{3}=
(a) \omega
(b) \omega^{2}
(c) 1
(d) -1

Answer:
1


11. \omega=\frac{1}{\omega^{2}}
(a) 1
(b) \omega^{3}
(c) \frac{1}{\omega}
(d) \frac{1}{\omega^{2}}

Answer: \frac{1}{\omega^{2}}
Explanation:
1.\omega \cdot \omega^{2}=1
Then
\omega=\frac{1}{\omega^{2}}


12. \omega^{2}=
(a) \omega
(b) \omega^{3}
(c) \frac{1}{\omega}
(d) \frac{1}{\omega^{2}}

Answer: \frac{1}{\omega}
Explanation:
1.\omega \cdot \omega^{2}=1
Then
\omega^{2}=\frac{1}{\omega}


13. \omega \cdot \omega^{2}=
(a) 1
(b) \omega^{3}
(c) \frac{1}{\omega}
(d) \frac{1}{\omega^{2}}

Answer:
1


14. (x+y)(x+\omega y)\left(x+\omega^{2} y\right)=
(a) \omega^{3}-\omega
(b) \omega^{3}
(c) x^{3}+y^{3}
(d) x^{3}-y^{3}

Answer:
Explanation:
(x+y)(x+\omega y)\left(x+\omega^{2} y\right)
=(x+y)\left(x^{2}+\omega^{2} x y+\omega x y+\omega^{3} y^{2}\right)
=(x+y)\left[x^{2}+x y\left(\omega^{2}+\omega\right)+\omega^{3} y^{2}\right]
As {\omega}^{2}+{\omega}=-1
=(x+y)\left[x^{2}+x y(-1)+(1)^{3} y^{2}\right]
=(x+y)\left(x^{2}-x y+y^{2}\right)
=x^{3}+y^{3}


15. Evaluate \omega^{15}
(a) \omega
(b) \omega^{2}
(c) 1
(d) -1

Answer:
Explanation:
\omega^{15}=\omega^{3 \times 5}
=\left(\omega^{3}\right)^{5}
=(1)^{5}
=1


16. Let \omega be one of the complex cubed roots of unity, then the value of 3+\omega+\omega^2 is
(a) 0
(b) 1
(c) 2
(d) 3

Answer: 2
Explanation:
3+\omega+\omega^2
3+(-1)
3-1
=2

Exercise # 2.3 MCQs

Sum and Product of Roots

In this portion of Chapter 2 Maths Class 10 includes the MCQs Sum of Roots and Product of Roots.

1. Sum of roots =
(a) \frac{-b}{a}
(b) \frac{c}{a}
(c) \omega
(d) \omega^{2}

Answer: \frac{-b}{a}


2. Product of roots =
(a) \frac{-b}{a}
(b) \frac{c}{a}
(c) \omega
(d) \omega^{2}

Answer: \frac{c}{a}


3. Sum of the roots of 2 x^{2}-3 x-4=0
(a) \frac{3}{2}
(b) -2
(c) \omega
(d) \omega^{2}

Answer: \frac{3}{2}
Explanation:
2 x^{2}-3 x-4=0
Compare it with a x^{2}+b x+c=0
Here a=2, b=-3, c=-4
Let \alpha \ and \ \beta be the roots of equation
Then sum of roots:
\alpha+\beta=\frac{-b}{a}
\alpha+\beta=\frac{-(-3)}{2}
\alpha+\beta=\frac{3}{2}


4. Products of the roots of 2 x^{2}-3 x-4=0
(a) \frac{3}{2}
(b) -2
(c) \omega
(d) \omega^{2}

Answer: -2
Explanation:
2 x^{2}-3 x-4=0
Compare it with a x^{2}+b x+c=0
Here a=2, b=-3, c=-4
Let \alpha \ and \ \beta be the roots of equation
Product of roots:
\alpha \cdot \beta=\frac{c}{a}
\alpha \cdot \beta=\frac{-4}{2}
\alpha \cdot \beta=-2

Exercise # 2.4 MCQs

In this Exercise of Chapter 2 Maths Class 10 includes the MCQs of Symmetric Functions of Roots of a Quadratic Equation.

Symmetric Functions of Roots

5. Let \alpha, \beta be the roots of a quadratic equation, then the expressions of the form of \alpha+\beta, \alpha \beta, \alpha^{2}+\beta^{2} are called the _________ of the roots of the quadratic equation.
(a) Solutions
(b) Roots
(c) Functions
(d) None of these

Answer: Functions


6. By _________ function of the roots of an equation, we mean that the function remains unchanged in values when the roots are interchanged.
(a) Solution
(b) Root
(c) Symmetric
(d) None of these

Answer: Symmetric


7. The functions \alpha+\beta, \alpha^{2}+\beta^{2}, \alpha^{3}+\beta^{3} are _________ function of \alpha \ and \ \beta
(a) Solution
(b) Root
(c) Symmetric
(d) None of these

Answer: Symmetric
Explanation:
By Symmetric function of the roots of an equation, we mean that the function remains unchanged in values when the roots are interchanged.


8. {\alpha}^{2}+{\beta}^{2}=
(a) (\alpha+\beta)^{2}
(b) (\alpha+\beta)^{2}-2 \alpha \beta
(c) (\alpha-\beta)^{2}-2 \alpha \beta
(d) None of these

Answer: (\alpha+\beta)^{2}-2 \alpha \beta
Explanation:
As
\alpha^{2}+\beta^{2}+2 \alpha \beta=(\alpha+\beta)^{2}
\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta
Then
\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta


9. \alpha^{3}+{\beta}^{3}=
(a) (\alpha+\beta)^{3}
(b) (\alpha+\beta)^{3}-3 \alpha \beta
(c) -3 \alpha \beta(\alpha+\beta)
(d) (\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)

Answer: (\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)
Explanation:
As
\alpha^{3}+\beta^{3}+3 \alpha\beta(\alpha+\beta)=(\alpha+\beta)^{3}
Then
\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)


10. \frac{1}{\alpha}+\frac{1}{\beta}=
(a) \frac{\alpha+\beta}{\alpha \beta}
(b) \beta+\alpha
(c) \beta \alpha
(d) \alpha+\beta

Answer: \frac{\alpha+\beta}{\alpha \beta}
Explanation:
\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}
\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}


11. If S and P be the sum and product of roots of a quadratic equation respectively, then the quadratic equation is
(a) x^2+Sx+P=0
(b) x^2-Sx+P=0
(c) x^2-Sx-P=0
(d) x^2+Sx-P=0

Answer: x^2-Sx+P=0
Explanation:
As a x^2+b x+c=0
Divide all terms by $a$
\frac{a x^2}{a}+\frac{b x}{a}+\frac{c}{a}=\frac{0}{a}
x^2+\frac{b x}{a}+\frac{c}{a}=0
Now we can write it as
x^2-\left(-\frac{b}{a}\right) x+\frac{c}{a}=0
As -\frac{b}{a}=\mathrm{S} \ and \ \frac{c}{a}=\mathrm{P}
Then
x^2-\mathrm{S} x+\mathrm{P}=0


12. Form a quadratic equation whose roots are 1+\sqrt{5}, 1-\sqrt{5}
(a) x^{2}-\sqrt{5} x+1=0
(b) x^{2}-S x+P=0
(c) x^{2}-2 x-4=0
(d) None of these

Answer: x^{2}-2 x-4=0
Explanation:
As 1+\sqrt{5} and 1-\sqrt{5} are the roots of required equation
Then sum of roots:
S=1+\sqrt{5}+1-\sqrt{5}
S=1+1+\sqrt{5}-\sqrt{5}
S=2
And product of roots:
P=(1+\sqrt{5})(1-\sqrt{5})
P=(1)^2-(\sqrt{5})^2
P=1-5
P=-4
As required equation is:
x^2-S x+P=0
Now
x^2-2 x+(-4)=0
x^2-2 x-4=0


13. (\alpha-\beta)^{2}=
(a) (\alpha+\beta)^{2}
(b) (\alpha+\beta)^{2}-4 \alpha \beta
(c) 2 \alpha \beta
(d) 4 \alpha \beta

Answer: (\alpha+\beta)^{2}-4 \alpha \beta
Explanation:
(\alpha-\beta)^{2}
=(\alpha+\beta)^{2}-4 \alpha \beta


14. ___________ division is the process of finding the quotient and remainder with less writing and fewer calculations.
(a) Synthetic
(b) Long
(c) No
(d) None of these

Answer: Synthetic


15. ___________ division is the shortcut of long division method and allows one to calculate without writing variables.
(a) Synthetic
(b) Long
(c) No
(d) None of these

Answer: Synthetic


16.___________ division can be used only when the divisor is a linear factor.
(a) Synthetic
(b) Long
(c) No
(d) None of these

Answer: Synthetic


17. More than one equation which are satisfied by the same values of the variables involved are called ___________ equations.
(a) Quadratic
(b) Linear
(c) Simultaneous
(d) None of these

Answer: Simultaneous


18. A system of Linear equation consists of two or more ___________ equations in the same variables.
(a) Quadratic
(b) Linear
(c) Function
(d) None of these

Answer: Linear


19. The sum of roots of an equation x^2-5kx++6k^2=0 is
(a) 5k
(b) -5k
(c) 6k^2
(d) None of these

Answer: 5k
Explanation:
x^2-5kx++6k^2=0
Compare it with a x^{2}+b x+c=0
Here a=1, b=-5k, c=6k^2
Let \alpha \ and \ \beta be the roots of equation
Then sum of roots:
\alpha+\beta=\frac{-b}{a}
\alpha+\beta=\frac{-(-5k)}{1}
\alpha+\beta=5k

Review Chapter 2 Mahts Class 10

(i). If the sum of roots of
(a+1)x^2+(2a+3)x+(3a+4)=0
is -1 , then roots is

O 0
O 1
O 2
O 3


2
Explanation:
(a+1)x^2+(2a+3)x+(3a+4)=0
a=a+1, b=2a+2, c=3a+4
As
Sum of roots =s=-1
Now
S=\frac{-b}{a}
-1=\frac{-(2a+3)}{a+1}
-1(a+1)=-2a-3
-a-1=-2a-3
-a+2a=-3+1
a=-2
Now
P=\frac{c}{a}
P=\frac{3a+4}{a+1}
Put a=-2
P=\frac{3(-2)+4}{-2+1}
P=\frac{-6+4}{-1}
P=\frac{-2}{-1}
P=2


(ii). The sum of the roots of a quadratic equation is 2 and the sum of the cubes of the roots is 98. The equation is
O x^2-2x-15=0
O x^2-2x+15=0
O x^2-4x+15=0
O None of these


x^2-2x-15=0
Explanation:
Let \alpha, \beta be the roots of quatratic eaquation
As we know that
S=\alpha+ \beta
P=\alpha \beta
Now we have
Sum of roots
S=\alpha+ \beta=2
and sum of cumbes of the roots \alpha^3+ \beta^3 =98
Now
(\alpha+ \beta)^3 -3(\alpha \beta)(\alpha+ \beta)=98
8 -6(\alpha \beta)=98
-6(\alpha \beta)=98-8
-6(\alpha \beta)=90
\alpha \beta=\frac{90}{-6}
\alpha \beta=-15
So
P=\alpha \beta=-15 For
As required equation is
x^2-Sx+P=0
So
x^2-2x+(-15)=0
x^2-2x-15=0


(iii). If a, b, c are positive real number, then both the roots of the equation a^2+bx+c=0 , are always
O real and positive
O real and negative
O rational and unequal
O none of these

None of these


(iv). If a and b are the roots of 4x^2-3x+7=0 then the value of \frac{1}{a}+ \frac{1}{b} is
O -\frac{3}{4}
O \frac{3}{7}
O -\frac{3}{7}
O \frac{4}{7}


\frac{3}{7}
Explanation:
4x^2-3x+7=0
a=4, b=-3, c=7
Let a and b be the roots of equation
Then sum of roots:
a+b=\frac{-b}{a}=\frac{-(-3)}{4}
a+b=\frac{3}{4}
and Product of roots:
ab=\frac{c}{a}=\frac{7}{4}
According to given condition
\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}
\frac{1}{a}+\frac{1}{b}=\frac{3/4}{7/4}
\frac{1}{a}+\frac{1}{b}=\frac{3}{7}

jawad khalil

jawad khalil

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