Skip to main content
Back to THEORY OF EQUATION

Your Progress

0 / 63 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • THEORY OF EQUATION

Theory of equation practice · Easy

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 1-30 of 63 questions
1 Easy
1 of 63
Analyzing +4 | -1
If x1,x2,x3,…,xnx_1, x_2, x_3, \ldots, x_n are the roots of the equation xn+ax+b=0x^n+a x+b=0, the value of (x1−x2)(x1−x3)(x1−x4)…(x1−xn)\left(x_1-x_2\right)\left(x_1-x_3\right)\left(x_1-x_4\right) \ldots\left(x_1-x_n\right) is
View answer and explanation
Correct option: C
n(x1)n−1+an\left(x_1\right)^{n-1}+a
Option C (n(x1)n−1+an(x_1)^{n-1}+a) is correct because it is the result of differentiating the polynomial and evaluating it at one of its roots.
2 Easy
2 of 63
Application +4 | -1
The polynomials P(x)=kx3+3x2−3P(x)=k x^3+3 x^2-3 and Q(x)=2x3−5x+kQ(x)=2 x^3-5 x+k, when divided by (x−4)(x-4) leave the same remainder, then kk is equal to:
View answer and explanation
Correct option: B
1
Option B (1) is correct because when the polynomials are divided by (x-4), setting the remainders equal and solving for k results in k=1.
3 Easy
3 of 63
Analyzing +4 | -1
If (x−1)3{\left( {x - 1} \right)^3} is a factor of x4+ax3+bx2+cx−1,{x^4} + a{x^3} + b{x^2} + cx - 1, then the other factor is
View answer and explanation
Correct option: B
x+1x + 1
Option B (x+1x + 1) is correct because by using the given factor and polynomial division or by finding the roots, we can determine the remaining factor.
4 Easy
4 of 63
Analyzing +4 | -1
The number of positive integral solutions of x4−y4=3789108x^4-y^4=3789108 is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the given equation has no positive integral solutions since the left side is always divisible by 8, while the right side is not.
5 Easy
5 of 63
Application +4 | -1
If α,β,γ\alpha ,\beta ,\gamma are the roots of x3−2x2+3x−4=0{x^3} - 2{x^2} + 3x - 4 = 0, then the value of α2β2+β2γ2+γ2α2{\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2} is
View answer and explanation
Correct option: A
−7- 7
Option A (−7- 7) is correct because it is the value of the expression calculated using Vieta's formulas.
6 Easy
6 of 63
Application +4 | -1
If the roots of the equation 4x3−12x2+11x+k=04{x^3} - 12{x^2} + 11x + k = 0 are in arithmetic progression, then kk is equal to
View answer and explanation
Correct option: A
−3- 3
Option D (None of these) is correct because the determinant simplifies to x2, which is not among the given options.
7 Easy
7 of 63
Application +4 | -1
The sum of all the real values of xx satisfying the equation 2(x−1)(x2+5x−50)=12^{(\mathrm{x}-1)\left(\mathrm{x}^{2}+5 \mathrm{x}-50\right)}=1 is :
View answer and explanation
Correct option: C
-4
Option C (-4) is correct because the sum of all real values of x that satisfy the equation is -4, which is obtained by solving the equation and summing the real roots.
8 Easy
8 of 63
Application +4 | -1
Let α,β\alpha, \beta be the roots of the quadratic equation x2+px+p3=0(p≠0)x^2+p x+p^3=0(p \neq 0). If (α,β)(\alpha, \beta) is a point on the parabola y2=xy^2=x, then the roots of the quadratic equation are
View answer and explanation
Correct option: A
4,−24,-2
Option A (4, -2) is correct because it satisfies both the quadratic equation and the parabola equation.
9 Easy
9 of 63
Application +4 | -1
If α,β,γ\alpha ,\beta ,\gamma are the roots of the equation x3+4x−1=0{x^3} + 4x - 1 = 0, then (α+β)−1+(β+γ)−1+(γ+α)−1{\left( {\alpha + \beta } \right)^{ - 1}} + {\left( {\beta + \gamma } \right)^{ - 1}} + {\left( {\gamma + \alpha } \right)^{ - 1}} is equal to
View answer and explanation
Correct option: C
−4- 4
Option C (-4) is correct because it is the result of simplifying the expression using Vieta's formulas for the sum and product of roots of the given cubic equation.
10 Easy
10 of 63
Analyzing +4 | -1
Number of real roots of the equation x+x−(1−x)=1\sqrt{x}+\sqrt{x-\sqrt{(1-x)}}=1 is
View answer and explanation
Correct option: B
1
The correct answer is B (1) because after simplification and squaring, only one real root satisfies the original equation.
11 Easy
11 of 63
Application +4 | -1
The equation log⁡3(3x−8)=2−x\log _3\left(3^x-8\right)=2-x has the solution
View answer and explanation
Correct option: B
x=2\mathrm{x}=2
Option B (x=2) is correct because it is the only solution that satisfies the given logarithmic equation after simplification and solving the resulting exponential equation.
12 Easy
12 of 63
Application +4 | -1
If α,β\alpha ,\beta and γ\gamma are the roots of x3+8=0,{x^3} + 8 = 0, then the equation whose roots are α2,β2{\alpha ^2},{\beta ^2} and γ2{\gamma ^2} is
View answer and explanation
Correct option: D
x3−64=0{x^3} - 64 = 0
The correct answer is D (x3−64=0{x^3} - 64 = 0) because it is derived by substituting x=yx = \sqrt{y} into the original equation and simplifying.
13 Easy
13 of 63
Analyzing +4 | -1
The real number kk for which the equation, 2x3+3x+k=02{x^3} + 3x + k = 0 has two distinct real roots in [0, 1]
View answer and explanation
Correct option: D
does not exist
Option D (does not exist) is correct because the derivative of the function is always positive, meaning the function is strictly increasing and can have at most one real root.
14 Easy
14 of 63
Application +4 | -1
If α,  β,γ\alpha ,\;\beta ,\gamma are the roots of x3+2x2−3x−1=0{x^3} + 2{x^2} - 3x - 1 = 0, then α−2+β−2+γ−2{\alpha ^{ - 2}} + {\beta ^{ - 2}} + {\gamma ^{ - 2}} is equal to
View answer and explanation
Correct option: B
13
Option B (13) is correct because by using Vieta's formulas and manipulating the expressions, we can find the value of the given expression.
15 Easy
15 of 63
Analyzing +4 | -1
Let α,β,γ\alpha, \beta, \gamma be the roots of the equation (x−a)(x−b)(x−c)=d,d≠0(x-a)(x-b)(x-c)=d, d \neq 0, the roots of the equation (x−α)(x−β)(x−γ)+d=0(x-\alpha)(x-\beta)(x-\gamma)+d=0 are
View answer and explanation
Correct option: C
a,b,ca, b, c
Option C (a,b,ca, b, c) is correct because by substituting the roots of the first equation into a modified second equation, we find that a,b,a, b, and cc are the roots.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
16 Easy
16 of 63
Analyzing +4 | -1
Maximum number of real solution for the equation axn+x2+bx+c=0a x^n+x^2+b x+c=0, where a,b,c∈Ra, b, c \in R and nn is an even positive number, is
View answer and explanation
Correct option: D
infinite
Option D (infinite) is correct because by choosing specific values for a, b, and c, the equation can have infinite solutions.
17 Easy
17 of 63
Application +4 | -1
If (x+1)(x+1) is factor of x3+kx2−3x+k+2x^3+k x^2-3 x+k+2, then kk is equal to:
View answer and explanation
Correct option: C
−2-2
Option C is correct because if (x+1) is a factor, then substituting x = -1 into the polynomial results in 0, which solves to k = -2.
18 Easy
18 of 63
Analyzing +4 | -1
The number of pairs (x,y)(x, y) which will satisfy the equation x2−xy+y2=4(x+y−4)x^2-x y+y^2=4(x+y-4), is
View answer and explanation
Correct option: A
1
Option A (1) is correct because by completing the square and analyzing the discriminant, we find only one pair (4,4) satisfies the equation.
19 Easy
19 of 63
Application +4 | -1
If α,β\alpha ,\beta are roots of ax2+bx+c=0,a{x^2} + bx + c = 0, then the equation ax2−bx(x−1)+c(x−1)2=0a{x^2} - bx\left( {x - 1} \right) + c{\left( {x - 1} \right)^2} = 0 has roots
View answer and explanation
Correct option: C
αα+1,ββ+1\frac{\alpha }{{\alpha + 1}},\frac{\beta }{{\beta + 1}}
Option C (αα+1,ββ+1\frac{\alpha}{{\alpha + 1}},\frac{\beta}{{\beta + 1}}) is correct because the roots of the transformed quadratic equation are related to the roots of the original equation in this specific way.
20 Easy
20 of 63
Application +4 | -1
The roots of an equation x3−9x2+14x+24=0x^3-9 x^2+14 x+24=0 are in the ratio 3:23: 2. The roots will be
View answer and explanation
Correct option: A
6,4,−16,4,-1
Option A (6, 4, -1) is correct because the roots are in the ratio 3:2 and their sum is 9, satisfying Vieta's formulas.
21 Easy
21 of 63
Analyzing +4 | -1
The condition that x3−px2+qx−r=0{x^3} - p{x^2} + qx - r = 0 may have two of its roots equal in magnitude but of opposite sign, is
View answer and explanation
Correct option: A
r=pqr = pq
Option A (r=pqr = pq) is correct because if two roots are equal in magnitude but opposite in sign, one root is pp, and substituting this into the equation gives r=pqr = pq.
22 Easy
22 of 63
Application +4 | -1
If x2−3x+2{x^2} - 3x + 2 be a factor of x4−px2+q,  then  (p,q){x^4} - p{x^2} + q,\;\mathrm{then}\;\left( {p,q} \right) is equal to
View answer and explanation
Correct option: D
(5,4)\left( {5,4} \right)
Option D ((5,4)) is correct because when x2−3x+2{x^2} - 3x + 2 is a factor of x4−px2+q{x^4} - p{x^2} + q, the roots of x2−3x+2=0{x^2} - 3x + 2 = 0 must also be roots of x4−px2+q=0{x^4} - p{x^2} + q = 0. Substituting these roots allows us to solve for p and q.
23 Easy
23 of 63
Application +4 | -1
If the roots of x3−3x2−6x+8=0{x^3} - 3{x^2} - 6x + 8 = 0 are in arithmetic progression, then the roots of the equation are
View answer and explanation
Correct option: C
−2,1,4- 2,1,4
Option C (−2,1,4-2, 1, 4) is correct because the sum of these roots is 3, which matches the coefficient of the x2 term in the given cubic equation.
24 Easy
24 of 63
Analyzing +4 | -1
If the two roots of the equation, (a−1)(x4+x2+1)+(a-1)\left(x^{4}+x^{2}+1\right)+ (a+1)(x2+x+1)2=0(a+1)\left(x^{2}+x+1\right)^{2}=0 are real and distinct, then the set of all values of ' aa ' is :
View answer and explanation
Correct option: B
(−12,0)∪(0,12)\left(-\frac{1}{2}, 0\right) \cup\left(0, \frac{1}{2}\right)
Option B ((−12,0)∪(0,12)\left(-\frac{1}{2}, 0\right) \cup\left(0, \frac{1}{2}\right)) is correct because by simplifying the given equation and analyzing the discriminant of the resulting quadratic, we can find the range of values for 'a' that satisfy the condition of having real and distinct roots.
25 Easy
25 of 63
Application +4 | -1
If the roots of x3−12x2+39x−28=0{x^3} - 12{x^2} + 39x - 28 = 0 are in A.P., then their common difference is
View answer and explanation
Correct option: C
±  3\pm \;3
Option C (±  3\pm \;3) is correct because the common difference is found by using the properties of Arithmetic Progression and solving the resulting equations.
26 Easy
26 of 63
Application +4 | -1
If α,β,γ,δ\alpha, \beta, \gamma, \delta are the roots of the equation x4+x3+x2x^4+x^3+x^2 +x+1=0+x+1=0, then α2021+β2021+γ2021+δ2021\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021} is equal to
View answer and explanation
Correct option: B
−1-1
Option A ((y2+x)4=C∣(y2+2x)3∣\left(y^2+x\right)^4=C\left|\left(y^2+2 x\right)^3\right|) is correct because it represents the general solution obtained after solving the given differential equation using appropriate substitutions and integration.
27 Easy
27 of 63
Application +4 | -1
If the sum of two of the roots of x3+px2−qx+r=0{x^3} + p{x^2} - qx + r = 0 is zero, then pqpq is equal to
View answer and explanation
Correct option: A
−r- r
Option A (−r-r) is correct because, given the sum of two roots is zero, pqpq simplifies to −r-r using Vieta's formulas.
28 Easy
28 of 63
Application +4 | -1
If one root of the equation ix2−2(1+i)x+2−i=0i x^2-2(1+i) x+2-i=0 is (3−i)(3-i), where i=−1i=\sqrt{-1}, the other root is
View answer and explanation
Correct option: D
−1−i-1-i
Option D (-1-i) is correct because by using the sum of roots property for quadratic equations, the other root can be found.
29 Easy
29 of 63
Application +4 | -1
If α1,  α2,  α3,  α4{\alpha _1},\;{\alpha _2},\;{\alpha _3},\;{\alpha _4} are the roots of the equation x4+(2−3)x2+2+3=0{x^4} + \left( {2 - \sqrt 3 } \right){x^2} + 2 + \sqrt 3 = 0, then the value of (1−α1)(1−α2)(1−α3)(1−α4)\left( {1 - {\alpha _1}} \right)\left( {1 - {\alpha _2}} \right)\left( {1 - {\alpha _3}} \right)\left( {1 - {\alpha _4}} \right) is
View answer and explanation
Correct option: D
5
Option D (5) is correct because using Vieta's formulas and substituting into the expression yields 5.
30 Easy
30 of 63
Application +4 | -1
If α,β,γ\alpha ,\beta ,\gamma are the roots of the equation 2x3−3x2+6x+1=02{x^3} - 3{x^2} + 6x + 1 = 0, then α2+β2+γ2{\alpha ^2} + {\beta ^2} + {\gamma ^2} is equal to
View answer and explanation
Correct option: A
−154- \frac{{15}}{4}
Option A (−154\frac{{ - 15}}{4}) is correct because it is the result of applying Vieta's formulas and algebraic manipulation to find the sum of the squares of the roots.
Revision notes and key takeaways

Review THEORY OF EQUATION in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

  • ✓ This set contains 63 single-correct-answer MCQs on THEORY OF EQUATION.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
Showing 1-30 of 63 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - THEORY OF EQUATION MCQs (All) 120 MCQs → ENGINEERING MATHEMATICS - THEORY OF EQUATION MCQs (Hard) 9 MCQs → ENGINEERING MATHEMATICS - THEORY OF EQUATION MCQs (Medium) 48 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs
📚 Switch Set (EASY)
Chapter Sets
ALL 120 MCQs EASY 63 MCQs HARD 9 MCQs MEDIUM 48 MCQs
0 /63
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →