Skip to main content
Back to QUADRATIC EQUATIONS

Your Progress

0 / 362 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Medium

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 61-90 of 362 questions
61 Medium
61 of 362
Analyzing +4 | -1
If roots of the equation ax2+bx+c=0;(a,b,c∈N)a{x^2} + bx + c = 0;\left( {a,b,c \in N} \right) are rational numbers, then which of the following cannot be true?
View answer and explanation
Correct option: B
All a,ba,b and cc are odd
Option B (All a,ba,b and cc are odd) cannot be true because it leads to a contradiction when considering the discriminant and the nature of rational roots.
62 Medium
62 of 362
Analyzing +4 | -1
The set of all the values of 'aa' for which the roots of the equation (a−2)x2−2ax+a=0(a-2) x^2-2 a x+a=0 lies in the interval (−2,1)(-2,1).
View answer and explanation
Correct option: A
a∈[0,89)∪{2}a \in\left[0, \frac{8}{9}\right) \cup\{2\}
The correct answer is A (a∈[0,89)∪{2}a \in\left[0, \frac{8}{9}\right) \cup\{2\}) because it represents the set of all 'a' values for which the roots of the given quadratic equation lie in the interval (-2, 1), considering different cases and conditions.
63 Medium
63 of 362
Analyzing +4 | -1
If the roots of the equation x2−2ax+a2+a−3=0{x^2} - 2ax + {a^2} + a - 3 = 0 are real and less than 3, then
View answer and explanation
Correct option: A
a<2a < 2
Option A (a<2a < 2) is correct because the roots being real and less than 3 imposes this constraint on the value of a.
64 Medium
64 of 362
Application +4 | -1
If x,y,z∈Rx, y, z \in \mathbf{R} then the least value of the expression E=3x2+5y2+4z2−6x+20y−8z−3E=3 x^2+5 y^2+4 z^2-6 x+20 y-8 z-3 is:
View answer and explanation
Correct option: B
-30
Option B (-30) is correct because completing the square for each variable allows us to rewrite the expression as a sum of squares plus a constant, and the minimum value occurs when the squares are zero.
65 Medium
65 of 362
Application +4 | -1
A real root of the equation log⁡4{log⁡2(x+8−x)}=0\log _4\left\{\log _2(\sqrt{x+8}-\sqrt{x})\right\}=0 is:
View answer and explanation
Correct option: A
1
Option A (1) is correct because substituting x=1 into the equation satisfies the condition.
66 Medium
66 of 362
Analyzing +4 | -1
If the equation, x2+bx+45=0(b∈R)x^{2}+b x+45=0(b \in R) has conjugate complex roots and they satisfy ∣z+1∣=210|z+1|=2 \sqrt{10}, then:
]
View answer and explanation
Correct option: A
b2−b=30b^{2}-b=30
Option A (b2−b=30b^{2}-b=30) is correct because by using the given condition ∣z+1∣=210|z+1|=2 \sqrt{10} and the properties of complex conjugate roots, we can find the value of b and then calculate b2−bb^{2}-b.
67 Medium
67 of 362
Application +4 | -1
The number of real solutions of the equation x2−3∣x∣+2=0x^{2}-3|x|+2=0 is
View answer and explanation
Correct option: C
4
Option C (4) is correct because solving the equation x2−3∣x∣+2=0x^{2}-3|x|+2=0 yields four real solutions: x=±1,±2x= \pm 1, \pm 2.
68 Medium
68 of 362
Application +4 | -1
Let a,b,ca, b, c be the lengths of the sides of a triangle (no two of them are equal) and k∈Rk \in R. If the roots of the equation x2+2(a+b+c)x+6k(ab+bc+ca)=0x^2+2(a+b+c) x+6 k(a b+b c+c a)=0 are real, then:
View answer and explanation
Correct option: A
k<23k<\frac{2}{3}
Option A (k<23k<\frac{2}{3}) is correct because the discriminant of the quadratic equation must be non-negative for real roots, leading to the inequality k<23k < \frac{2}{3}.
69 Medium
69 of 362
Analyzing +4 | -1
The values of xx satisfying ∣x2+4x+3∣+(2x+5)=0\left| {{x^2} + 4x + 3} \right| + \left( {2x + 5} \right) = 0 are
View answer and explanation
Correct option: A
−4,  −1−3- 4,\; - 1 - \sqrt 3
Option A (−4,  −1−3- 4,\; - 1 - \sqrt 3) is correct because it involves solving the equation by considering cases based on the absolute value and checking the validity of the solutions.
70 Medium
70 of 362
Application +4 | -1
If log⁡2x+log⁡x2=103=log⁡2y+log⁡y2\log _2 x+\log _x 2=\frac{10}{3}=\log _2 y+\log _y 2 and x≠yx \neq y, then (x+y)=?(x+y)=?
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because solving the logarithmic equation leads to x=8x=8 and y=21/3y=2^{1/3}, so x+y=8+21/3x+y = 8 + 2^{1/3}.
71 Medium
71 of 362
Application +4 | -1
If a,ba, b and cc are real numbers such that a2+b2+c2=1a^2+b^2+c^2=1, then ab+bc+caa b+b c+c a lies in the interval
View answer and explanation
Correct option: B
[−12,1]\left[-\frac{1}{2}, 1\right]
The correct answer is B ([−12,1\frac{-1}{2}, 1]) because using (a+b+c)2≥0(a+b+c)^2 \geq 0 and (a−b)2+(b−c)2+(c−a)2≥0(a-b)^2+(b-c)^2+(c-a)^2 \geq 0 allows us to find the interval for ab+bc+caab+bc+ca.
72 Medium
72 of 362
Application +4 | -1
The value(s) of yy for which the equation 4sin⁡x+3cos⁡x=y2−6y+144 \sin x+3 \cos x=y^2-6 y+14 has a real solution, is (are)
View answer and explanation
Correct option: A
3
Option A (3) is correct because the equation 4sin⁡x+3cos⁡x=y2−6y+144 \sin x + 3 \cos x = y^2 - 6y + 14 has a real solution only when y=3y=3, which makes the right-hand side equal to 5, the maximum value of the left-hand side.
73 Medium
73 of 362
Application +4 | -1
Let f(x)f\left( x \right) be a quadratic expression which is positive for all real xx and g(x)=f(x)+f′(x)+f′′(x),g\left( x \right) = f\left( x \right) + f'\left( x \right) + f''\left( x \right), then for any real xx,
View answer and explanation
Correct option: B
g(x)>0g\left( x \right) > 0
Option B (g(x)>0g(x) > 0) is correct because if f(x) is always positive, then g(x), which is a combination of f(x) and its derivatives, is also always positive.
74 Medium
74 of 362
Application +4 | -1
The solution of the inequation 4−x+0.5−7.2−x<4,x∈R{4^{ - x + 0.5}} - {7.2^{ - x}} < 4,x \in R IS
View answer and explanation
Correct option: A
(−2,∞)\left( { - 2,\infty } \right)
The correct answer is A ((−2,∞)\left( { - 2,\infty } \right)) because by substituting 2−x=t2^{-x} = t, the inequality can be solved for t, and then back-substituted to find the range of x.
75 Medium
75 of 362
Application +4 | -1
If x2−x−2x^2-x-2 is a factor of x4−px2−q=0x^4-p x^2-q=0, then (p+q)(p+q) equals
View answer and explanation
Correct option: B
1
The correct answer is B (1) because substituting the roots of x2−x−2x^2-x-2 into x4−px2−q=0x^4-px^2-q=0 allows us to solve for p+qp+q.

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
76 Medium
76 of 362
Application +4 | -1
If xx be real, then the minimum value of x2−8x+17x^2-8 x+17 is:
View answer and explanation
Correct option: C
1
Option C (1) is correct because the minimum value of the quadratic expression x2−8x+17x^2-8 x+17 can be found by completing the square or using the formula 4ac−b24a\frac{{4ac-b^2}}{{4a}}.
77 Medium
77 of 362
Application +4 | -1
The greatest negative integer satisfying x2−4x−77<0{x^2} - 4x - 77 < 0 and x2>4,{x^2} > 4, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the greatest negative integer satisfying both inequalities is -3, which is not among the options.
78 Medium
78 of 362
Application +4 | -1
If the roots of the equation x2−5x+16=0x^2-5 x+16=0 are α,β\alpha, \beta and the roots of equation x2+px+q=0x^2+p x+q=0 are α2+β2,αβ/2\alpha^2+\beta^2, \alpha \beta / 2, then
View answer and explanation
Correct option: B
p=−1,q=−56p=-1, q=-56
Option D (p=−1,q=56p=-1, q=56) is correct because by using Vieta's formulas and the relationships between the roots of the two equations, we can find the values of pp and qq.
79 Medium
79 of 362
Analyzing +4 | -1
The equations x3+5x2+px+q=0x^3+5 x^2+p x+q=0 and x3+7x2+px+r=0x^3+7 x^2+p x+r=0 have two roots in common. If their third roots are γ1\gamma_1 and γ2\gamma_2 respectively, then the ordered pair (γ1,γ2)\left(\gamma_1, \gamma_2\right) is
View answer and explanation
Correct option: B
(−5,−7)(-5,-7)
Option B ((-5,-7)) is correct because the sum of the roots is -5 and -7 respectively, and the two common roots sum to zero.
80 Medium
80 of 362
Analyzing +4 | -1
The value of xx in the given equation 4x−3x−12=3x+12−22x−14^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2 x-1} is:
View answer and explanation
Correct option: B
32\frac{3}{2}
Option B (3/2) is the correct answer because it satisfies the given exponential equation after simplification and taking logarithms.
81 Medium
81 of 362
Analyzing +4 | -1
If the larger root of equation x2+(2−a2)x+(1−a2)=0x^2+\left(2-a^2\right) x+\left(1-a^2\right)=0 is less than both the roots of the equation x2−(a2+4a+1)x+a2+4a=0x^2-\left(a^2+4 a+1\right) x+a^2+4 a=0, then the range of a,a, is
View answer and explanation
Correct option: B
(−14,2)\left(-\frac{1}{4}, \sqrt{2}\right)
Option A (sin⁡−1(tan⁡α2⋅∣cot⁡α∣)\sin ^{-1}\left(\tan \frac{\alpha}{2} \cdot|\cot \alpha|\right)) is correct because it is derived by finding the angle between the vector product of two unit vectors and the plane containing the other two unit vectors.
82 Medium
82 of 362
Analyzing +4 | -1
For real θ\theta, value of sin⁡2θ+cos⁡4θ\sin ^2 \theta+\cos ^4 \theta lies in the interval
View answer and explanation
Correct option: B
[34,1]\left[\frac{3}{4}, 1\right]
Option B ([3/4, 1]) is correct because the expression \sin2 θ + \cos4 θ lies within this interval, as determined by its maximum and minimum values.
83 Medium
83 of 362
Application +4 | -1
If roots of the equation (a−b)x2+(c−a)x+(b−c)=0\left( {a - b} \right){x^2} + \left( {c - a} \right)x + \left( {b - c} \right) = 0 are equal, then a,b,ca,b,c are in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because the condition for equal roots leads to the relationship a+c=2ba + c = 2b, which defines an arithmetic progression.
84 Medium
84 of 362
Application +4 | -1
If pp and qq are roots of the quadratic equation x2+mx+m2+a=0,{x^2} + mx + {m^2} + a = 0, then the value of p2+q2+pq{p^2} + {q^2} + pq is
View answer and explanation
Correct option: C
−a- a
Option D (2) is correct because by substituting the given value of z and using the properties of logarithms and exponentials, we can find that cos z = 2.
85 Medium
85 of 362
Application +4 | -1
Let the two numbers have arithmetic mean 9 and geometric mean 4. Then, these numbers are the roots of the quadratic equation
View answer and explanation
Correct option: B
x2−18x+16=0{x^2} - 18x + 16 = 0
Option B (x2−18x+16=0{x^2} - 18x + 16 = 0) is correct because the sum of the roots is 18 (twice the arithmetic mean) and the product of the roots is 16 (the square of the geometric mean).
86 Medium
86 of 362
Application +4 | -1
Let for a≠a1≠0\mathrm{a} \neq a_{1} \neq 0,

f(x)=ax2+bx+c,g(x)=a1x2+b1x+c1f(x)=a x^{2}+b x+c, g(x)=a_{1} x^{2}+b_{1} x+c_{1}

।and p(x)=f(x)−g(x)p(x)=f(x)-g(x).

If p(x)=0p(x)=0 only for x=−1x=-1 and p(−2)=2p(-2)=2, then the value of pp (b) is :
View answer and explanation
Correct option: D
18
Option D (18) is correct because solving the given conditions for p(x) and evaluating p(2) yields 18.
87 Medium
87 of 362
Analyzing +4 | -1
If one of the roots of the equation x2+ax+b=0x^2+a x+b=0 and x2+x^2+ bx+a=0b x+a=0 is coincident. Then the numerical value of (a+b)(a+b) is:
View answer and explanation
Correct option: B
-1
Option B (-1) is correct because by equating the coincident root in both equations, we can derive that a+b = -1.
88 Medium
88 of 362
Analyzing +4 | -1
Suppose the quadratic equations x2+px+q=0{x^2} + px + q = 0 and x2+rx+s=0{x^2} + rx + s = 0 are such that p,  q,  r,  sp,\;q,\;r,\;s are real and pr=2(q+s)pr = 2\left( {q + s} \right). Then
View answer and explanation
Correct option: B
At least one equation always has real roots
Option B (At least one equation always has real roots) is correct because the sum of the discriminants is non-negative, implying at least one discriminant is non-negative.
89 Medium
89 of 362
Application +4 | -1
If log⁡2(ax2+x+a)≥1,∀x∈R\log _2\left(a x^2+x+a\right) \geq 1, \forall x \in R, then exhaustive set of values of aa is
View answer and explanation
Correct option: D
(1+52,∞)\left(1+\frac{\sqrt{5}}{2}, \infty\right)
The correct answer is D (1+52,∞)\left(1+\frac{{\sqrt{{5}}}}{{2}}, \infty\right) because it satisfies the condition that the quadratic expression is always greater than or equal to zero for all real x.
90 Medium
90 of 362
Analyzing +4 | -1
Let pp and qq be real numbers such that p≠0,p3≠−qp \neq 0, p^3 \neq-q. If α\alpha and β\beta are non-zero complex numbers satisfying α+β=−p\alpha+\beta=-p and α3+β3=q\alpha^3+\beta^3=q, a quadratic equation having αβ\frac{\alpha}{\beta} and βα\frac{\beta}{\alpha} as its roots, is
View answer and explanation
Correct option: B
(p3+q)x2−(p3−2q)x+(p3+q)=0\left(p^3+q\right) x^2-\left(p^3-2 q\right) x+\left(p^3+q\right)=0
Option B ((p3+q)x2−(p3−2q)x+(p3+q)=0\left(p^3+q\right) x^2-\left(p^3-2 q\right) x+\left(p^3+q\right)=0) is correct because it is derived by finding the sum and product of the roots αβ\frac{\alpha}{\beta} and βα\frac{\beta}{\alpha} in terms of pp and qq, and then constructing the quadratic equation.
Revision notes and key takeaways

Review QUADRATIC EQUATIONS in your textbook or class notes, then use this set to check your understanding.

Showing 61-90 of 362 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - QUADRATIC EQUATIONS MCQs (All) 914 MCQs → ENGINEERING MATHEMATICS - QUADRATIC EQUATIONS MCQs (Easy) 528 MCQs → ENGINEERING MATHEMATICS - QUADRATIC EQUATIONS MCQs (Hard) 24 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 (MEDIUM)
Chapter Sets
ALL 914 MCQs EASY 528 MCQs HARD 24 MCQs MEDIUM 362 MCQs
0 /362
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 →