Skip to main content
Back to QUADRATIC EQUATIONS

Your Progress

0 / 528 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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 331-360 of 528 questions

Preparing practice and saved progress…

331 Easy
331 of 528
Analyzing +4 | -1
The solution set of the inequation ∣2x−1x−1∣>2,\left| {\frac{{2x - 1}}{{x - 1}}} \right| > 2, is
View answer and explanation
Correct option: A
(3/4  ,1)∪(1,∞)\left( {3/4\;,1} \right) \cup \left( {1,\infty } \right)
Option A (3/4  ,1)∪(1,∞)\left( {3/4\;,1} \right) \cup \left( {1,\infty } \right) is the correct solution set obtained by solving the inequality by considering both positive and negative cases of the absolute value.
332 Easy
332 of 528
Application +4 | -1
(log⁡5x)2+(log⁡5x)<2,{\left( {{{\log }_5}x} \right)^2} + \left( {{{\log }_5}x} \right) < 2, then xx belongs to the interval
View answer and explanation
Correct option: A
(1/25  ,5)\left( {1/25\;,5} \right)
Option A ((1/25  ,5)\left( {1/25\;,5} \right)) is correct because it represents the interval to which x belongs after solving the logarithmic inequality.
333 Easy
333 of 528
Analyzing +4 | -1
If a,  ba,\;b and cc are in geometric progression and the roots of the equation ax2+2bx+c=0a{x^2} + 2bx + c = 0 are α\alpha and β\beta and those of cx2+2bx+a=0c{x^2} + 2bx + a = 0 are γ\gamma and δ\delta, then
View answer and explanation
Correct option: C
aα=aβ=cγ=cδa\alpha = a\beta = c\gamma = c\delta
Option D (aα=aβ=cγ=cδa\alpha = a\beta = c\gamma = c\delta) is correct because the roots of both quadratic equations are equal and related by the common ratio of the geometric progression.
334 Easy
334 of 528
Application +4 | -1
The sum of all real roots of the equation ∣x−2∣2+∣x−2∣−2=0|x-2|^2+|x-2|-2=0 is
View answer and explanation
Correct option: B
4
Option B (4) is correct because by substituting t=∣x−2∣t = |x-2|, we find the real roots x=1x=1 and x=3x=3, and their sum is 4.
335 Easy
335 of 528
Application +4 | -1
If ecos⁡x−e−cos⁡x=4,{e^{\cos x}} - {e^{ - \cos x}} = 4, then the value of cos⁡x\cos x is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the derived value of cos⁡x\cos x is greater than 1, which is not possible for the cosine function.
336 Easy
336 of 528
Application +4 | -1
If α,β\alpha, \beta are roots of 9x2−11x+1=09 x^2-11 x+1=0 then value of 1(9α−11)2+(11β−19)\frac{1}{(9 \alpha-11)^2}+\left(\frac{11 \beta-1}{9}\right) is-
View answer and explanation
Correct option: D
10381\frac{103}{81}
Option D (10381\frac{103}{81}) is correct because by using Vieta's formulas and the given quadratic equation, the expression simplifies to α2+β2=10381\alpha^2 + \beta^2 = \frac{103}{81}.
337 Easy
337 of 528
Application +4 | -1
If α+β=4\alpha + \beta = 4 and α3+β3=44{\alpha ^3} + {\beta ^3} = 44, then α,  β\alpha ,\;\beta are the roots of the equation
View answer and explanation
Correct option: D
3x2−12x+5=03{x^2} - 12x + 5 = 0
Option D (3x² - 12x + 5 = 0) is correct because using the given conditions to find the sum and product of the roots leads to this quadratic equation.
338 Easy
338 of 528
Application +4 | -1
If a,c≠0a,c \ne 0 and α,β\alpha ,\beta are the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0, then the quadratic equation with 1/α1/\alpha and 1/β1/\beta as its root is
View answer and explanation
Correct option: B
cx2+bx+a=0c{x^2} + bx + a = 0
Option B (cx2+bx+a=0c{x^2} + bx + a = 0) is correct because it represents the quadratic equation with roots that are reciprocals of the original roots.
339 Easy
339 of 528
Application +4 | -1
x+4x−3<2\frac{x+4}{x-3} < 2 is satisfied when xx satisfies
View answer and explanation
Correct option: A
(−∞,3)∪(10,∞)(-\infty, 3) \cup(10, \infty)
Option A ((−∞,3)∪(10,∞))((-\infty, 3) \cup (10, \infty)) is correct because it represents the set of xx values that satisfy the given inequality.
340 Easy
340 of 528
Application +4 | -1
If one root of the quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 is equal to nth power of the other root, then the value of (acn)1/(n+1)+(anc)1/(n+1){\left( {a{c^n}} \right)^{1/\left( {n + 1} \right)}} + {\left( {{a^n}c} \right)^{1/\left( {n + 1} \right)}} is equal to
View answer and explanation
Correct option: B
−b- b
Option B (−b-b) is correct because it is the result of eliminating α from the sum and product of the roots, and then simplifying the expression.
341 Easy
341 of 528
Application +4 | -1
Let α,β\alpha ,\beta be the roots of x2+bx+1=0.{x^2} + bx + 1 = 0. Then the equation whose roots are −(α+1β)- \left( {\alpha + \frac{1}{\beta }} \right) and −(β+1α),- \left( {\beta + \frac{1}{\alpha }} \right), is
View answer and explanation
Correct option: C
x2−2    b  x+4=0{x^2} - 2\;\;b\;x + 4 = 0
Option C (x2−2    b  x+4=0{x^2} - 2\;\;b\;x + 4 = 0) is correct because it is formed by finding the sum and product of the new roots and using them to construct the quadratic equation.
342 Easy
342 of 528
Application +4 | -1
Suppose that two persons AA and BB solve the equation x2+ax+b=0{x^2} + ax + b = 0. While solving AA commits a mistake in the coefficient of xx was taken as 15 in place of −9- 9 and finds the roots as −7- 7 and −2- 2. Then the equation is
View answer and explanation
Correct option: B
x2−9x+14=0{x^2} - 9x + 14 = 0
Option B (x2−9x+14=0{x^2} - 9x + 14 = 0) is correct because it uses the correct coefficient of the constant term (product of roots) obtained from the incorrect equation solved by person A.
343 Easy
343 of 528
Application +4 | -1
If the equations x2+ax+b=0x^{2}+a x+b=0 and x2+a′x+b′=0x^{2}+a^{\prime} x+b^{\prime}=0 have a common root, then this common root is equal to
View answer and explanation
Correct option: D
(b−b′)(a′−a)\frac{\left(b-b^{\prime}\right)}{\left(a^{\prime}-a\right)}
Option D ((b−b′)(a′−a)\frac{{\left(b-b^{\prime}\right)}}{{\left(a^{\prime}-a\right)}}) is correct because by subtracting the two equations, the common root can be isolated and solved for.
344 Easy
344 of 528
Analyzing +4 | -1
The value of 'kk' for which one of the roots of x2−x+3  k=0,{x^2} - x + 3\;k = 0, is double of one of the roots of x2−x+k=0{x^2} - x + k = 0 is
View answer and explanation
Correct option: B
−2- 2
Option B (−2- 2) is correct because by setting up the relationship between the roots of the two quadratic equations and solving the resulting equations, we find that k = -2.
345 Easy
345 of 528
Application +4 | -1
If α,β\alpha, \beta are roots of x2−3x+1=0x^2-3 x+1=0, then the equation whose roots are 1α−2,1β−2\frac{1}{\alpha-2}, \frac{1}{\beta-2} is:
View answer and explanation
Correct option: C
x2−x−1=0 dx^2-x-1=0 \mathrm{~d}
Option C (x2−x−1=0 dx^2-x-1=0 \mathrm{~d}) is correct because it is obtained by finding the sum and product of the new roots (1α−2\frac{1}{\alpha-2} and 1β−2\frac{1}{\beta-2}) and constructing the quadratic equation.
346 Easy
346 of 528
Application +4 | -1
The value of
x=4+15+14+15+14+...∞x = 4 + \frac{1}{{5 + \frac{1}{{4 + \frac{1}{{5 + \frac{1}{{4 + ...\infty }}}}}}}}
View answer and explanation
Correct option: A
2+25302 + \frac{2}{5}\sqrt {30}
Option A (2+25302 + \frac{2}{5}\sqrt {30}) is correct because it is the positive solution to the quadratic equation derived from the continued fraction.
347 Easy
347 of 528
Analyzing +4 | -1
If a<b<c<da < b < c < d, then the roots of the equation (x−a)(x−c)+2(x−b)(x−d)=0\left( {x - a} \right)\left( {x - c} \right) + 2\left( {x - b} \right)\left( {x - d} \right) = 0 are
View answer and explanation
Correct option: A
Real and distinct
Option A (Real and distinct) is correct because the discriminant of the given equation is positive, indicating that the roots are real and distinct.
348 Easy
348 of 528
Application +4 | -1
Two students while solving a quadratic equation in xx, one copied the constant term incorrectly and got the roots 3 and 2. The other copied the constant term coefficient of x2{x^2} correctly as −6- 6 and 1 respectively the correct roots are
View answer and explanation
Correct option: D
6,  −16,\; - 1
Option D (6,  −16,\; - 1) is correct because by using the information about the incorrect roots, we can determine the correct sum and product of the roots, leading to the correct roots of 6 and -1.
349 Easy
349 of 528
Application +4 | -1
The sum of all values of xx, so that 16(x2+3x−1)=8(x2+3x+2)16^{\left(x^2+3 x-1\right)}=8^{\left(x^2+3 x+2\right)}, is
View answer and explanation
Correct option: C
-3
Option C, -3, is correct because solving the exponential equation 16(x2+3x−1)=8(x2+3x+2)16^{(x^2+3x-1)} = 8^{(x^2+3x+2)} leads to a quadratic equation whose roots sum to -3.
350 Easy
350 of 528
Application +4 | -1
If α\alpha and β\beta are the roots of the equation x2+px+q=0x^2+p x+q=0, then the value of (ωα+ω2β)(ω2α+ωβ)\left(\omega \alpha+\omega^2 \beta\right)\left(\omega^2 \alpha+\omega \beta\right), where ω\omega is an imaginary cube root of unity, is
View answer and explanation
Correct option: B
p2−3qp^2-3 q
Option B (p2−3qp^2-3 q) is correct because it is the simplified expression of (ωα+ω2β)(ω2α+ωβ)(\omega \alpha+\omega^2 \beta)(\omega^2 \alpha+\omega \beta) using the properties of cube roots of unity and Vieta's formulas.
351 Easy
351 of 528
Application +4 | -1
The value of x=2+2+2+……x=\sqrt{2+\sqrt{2+\sqrt{2+\ldots \ldots}}} is;
View answer and explanation
Correct option: B
2
Option B (2) is correct because the value is found by setting up an equation x = sqrt(2+x) and solving the quadratic equation.
352 Easy
352 of 528
Analyzing +4 | -1
If α\alpha and β\beta are the distinct roots of the equation x2+(3)1/4x+31/2=0{x^2} + {(3)^{1/4}}x + {3^{1/2}} = 0 then the value of α96(α12−1)+β96(β12−1){\alpha ^{96}}\left( {{\alpha ^{12}} - 1} \right) + {\beta ^{96}}\left( {{\beta ^{12}} - 1} \right) is equal to
View answer and explanation
Correct option: C
52×32452 \times {3^{24}}
The correct answer is C (52×32452 \times {3^{24}}) because it results from simplifying the expression after substituting the roots of the given quadratic equation.
353 Easy
353 of 528
Application +4 | -1
If the equation 5x2−10x+log⁡1/5a=05 x^{2}-10 x+\log _{1 / 5} a=0 has real roots, then the minimum value of aa is
View answer and explanation
Correct option: A
155\frac{1}{5^{5}}
Option A (155\frac{{1}}{{5^{5}}}) is correct because the equation having real roots implies that the discriminant is non-negative, leading to the minimum value of aa.
354 Easy
354 of 528
Application +4 | -1
Let f(x)=∫1x(2−t2)dtf(x)=\int_1^x \sqrt{\left(2-t^2\right)} d t, the real roots of the equation x2−f′(x)=0x^2-f^{\prime}(x)=0 are
View answer and explanation
Correct option: A
±1\pm 1
Option A ±1\pm 1 is correct because these are the real roots obtained by solving the equation x2−f′(x)=0x^2 - f'(x) = 0 after finding f′(x)f'(x).
355 Easy
355 of 528
Application +4 | -1
If a,b∈R,a≠0a, b \in R, a \neq 0 and the quadratic equation ax2−bx+1=0a x^{2}-b x+1=0 has imaginary roots, then a+b+1a+b+1 is
View answer and explanation
Correct option: A
positive
Option A (positive) is correct because the quadratic equation having imaginary roots implies the function is always positive, and therefore f(-1) is positive.
356 Easy
356 of 528
Application +4 | -1
The product of the roots of the equation xlog⁡10x=104x3x^{\log _{10} x}=\frac{10^4}{x^3}, is
View answer and explanation
Correct option: C
11000\frac{1}{1000}
Option C (11000\frac{{1}}{{1000}}) is correct because it represents the product of the roots of the given logarithmic equation.
357 Easy
357 of 528
Application +4 | -1
If α\alpha and β\beta are the roots of the equation ax2+bx+c=0a x^2+b x+c=0, then roots of

ax2−bx(x−1)+c(x−1)2=0a x^2-b x(x-1)+c(x-1)^2=0 (1)

are
View answer and explanation
Correct option: C
α1+α,β1+β\frac{\alpha}{1+\alpha}, \frac{\beta}{1+\beta}
Option C (α1+α,β1+β\frac{\alpha}{1+\alpha}, \frac{\beta}{1+\beta}) is correct because it is derived by substituting −xx−1-\frac{x}{x-1} into the original equation and solving for x.
358 Easy
358 of 528
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 by solving the logarithmic equation step-by-step, we find that the real root is x=1x=1.
359 Easy
359 of 528
Application +4 | -1
The number of integral values of mm for which the equation (1+m2)x2−2(1+3m)x+(1+8m)=0,(1 + {m^2}){x^2} - 2(1 + 3m)x + (1 + 8m) = 0, has no real root is
View answer and explanation
Correct option: B
infinitely many
Option B (infinitely many) is correct because the discriminant of the quadratic equation must be negative, leading to the condition m > 0, which has infinitely many solutions.
360 Easy
360 of 528
Analyzing +4 | -1
The number of integral solutions of x+2x2+1>12\frac{{x + 2}}{{{x^2} + 1}} > \frac{1}{2} is
View answer and explanation
Correct option: C
3
Option C (3) is correct because there are three integers within the solution interval of the inequality.
Revision notes and key takeaways

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

Showing 331-360 of 528 questions

Your practice

0 / 528 answered · 528 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

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.