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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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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.

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Showing 31-60 of 914 questions
31 Medium
31 of 914
Application +4 | -1
The number of real values of xx satisfying the equation is

2(x2+1x2)−9(x+1x)+14=0   …(1)2\left(x^2+\frac{1}{x^2}\right)-9\left(x+\frac{1}{x}\right)+14=0\,\,\,…(1)
is
View answer and explanation
Correct option: C
3
Option C (3) is correct because there are three real values of x that satisfy the given equation after substitution and simplification.
32 Medium
32 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of the equation 2x2−3x−6=02 x^2-3 x-6=0, then the equation whose roots are α2+2\alpha^2+2 and β2+2\beta^2+2 is
View answer and explanation
Correct option: C
4x2−49x+118=04 x^2-49 x+118=0
Option C (4x2−49x+118=04 x^2-49 x+118=0) is correct because it represents the quadratic equation with roots α2+2\alpha^2+2 and β2+2\beta^2+2, derived from the original equation's roots α\alpha and β\beta.
33 Medium
33 of 914
Application +4 | -1
If one real root of the quadratic equation 81x2+kx+256=081{x^2} + kx + 256 = 0 is cube of the other root, then a value of k is
View answer and explanation
Correct option: D
-300
Option D (-300) is correct because it's the value of k derived from the relationship between the roots of the quadratic equation.
34 Easy
34 of 914
Application +4 | -1
If the sum of the roots of the equation (a+1)x2+(2a+3)x+(3a+4)=0\left( {a + 1} \right){x^2} + \left( {2a + 3} \right)x + \left( {3a + 4} \right) = 0 is −1- 1, then the product of the roots is
View answer and explanation
Correct option: C
2
Option C (2) is correct because by using the given information about the sum of the roots and Vieta's formulas, we can find the value of 'a' and then calculate the product of the roots.
35 Medium
35 of 914
Analyzing +4 | -1
The solution set for (2+2)x+(2−2)x=2⋅2x/4(\sqrt{2+\sqrt{2}})^x+(\sqrt{2-\sqrt{2}})^x=2 \cdot 2^{x / 4} is
View answer and explanation
Correct option: C
{0}\{0\}
Option C ({0}) is correct because x=0 is the only solution that satisfies the given equation, derived from the AM-GM inequality.
36 Easy
36 of 914
Application +4 | -1
The number real of roots of x−3(x2+7x+10)=0\sqrt{x-3}\left(x^2+7 x+10\right)=0, where x∈Rx \in \mathbf{R} is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the square root term requires x to be greater than or equal to 3, and only x=3 satisfies the equation.
37 Easy
37 of 914
Application +4 | -1
If one of the roots of the equation x2+bx+3=0{x^2} + bx + 3 = 0 is thrice the other, then bb is equal to
View answer and explanation
Correct option: D
±4\pm 4
Option D (±4\pm 4) is correct because by setting up the relationship between the roots and using Vieta's formulas, we can solve for the value of b.
38 Easy
38 of 914
Application +4 | -1
If the line x−23=y+15=z−2+2\frac{x-2}{3}=\frac{y+1}{5}=\frac{z-2}{+2} lies in the plane αx+2y+z−β\alpha x+2 y+z-\beta =0=0, then (α,β)(\alpha, \beta) equals
View answer and explanation
Correct option: C
(−4,−8)(-4,-8)
Option C ((-4,-8)) is correct because substituting the point (2,-1,2) into the plane equation and using the direction ratios of the line to satisfy the plane condition yields α = -4 and β = -8.
39 Easy
39 of 914
Application +4 | -1
if one of the roots of the equation x2+(1−3i)x−2(1+i)=0{x^2} + \left( {1 - 3i} \right)x - 2\left( {1 + i} \right) = 0 is −1+i,- 1 + i, then the other root is
View answer and explanation
Correct option: D
2i2i
Option D (2i2i) is correct because it is the other root of the quadratic equation, found using the sum of roots property.
40 Easy
40 of 914
Application +4 | -1
If xx is real, then the least value of the expression (ax2+bx+c),a>0\left(a x^2+b x+c\right), a>0 is
View answer and explanation
Correct option: C
4ac−b24a\frac{4 a c-b^2}{4 a}
Option C (4ac−b24a\frac{{4 a c-b^2}}{{4 a}}) is correct because it represents the minimum value of the quadratic expression, derived by completing the square.
41 Easy
41 of 914
Analyzing +4 | -1
Let SS denote the set of all values of SS for which the equation 2x2−2(2a+1)x+a(a+1)=02{x^2} - 2\left( {2a + 1} \right)x + a\left( {a + 1} \right) = 0 has one root less than aa and other root greater than aa, then SS equals
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the correct range for 'a' is a < -1 or a > 0, which is not represented by any of the given options.
42 Easy
42 of 914
Application +4 | -1
If α,β\alpha ,\beta are roots of x2+px−q=0{x^2} + px - q = 0 and γ,δ\gamma ,\delta are root of x2+px+r=0,{x^2} + px + r = 0, then the value of (α−γ)(α−δ)\left( {\alpha - \gamma } \right)\left( {\alpha - \delta } \right) is
View answer and explanation
Correct option: D
q+rq + r
Option D (q+rq + r) is correct because substituting the roots into the given equations and simplifying leads to this result.
43 Medium
43 of 914
Analyzing +4 | -1
It is given that P(x)=x3+ax2+bx+cP(x)=x^3+a x^2+b x+c has three distinct positive integral roots and P(34)=33P(34)=33. Let Q(x)=x2−2x+34Q(x)=x^2-2 x+34. If P(Q(x))P(Q(x)) has no real roots then cc is equal to
View answer and explanation
Correct option: D
-23529
Option D (-23529) is correct because by factoring P(x) and using P(34) = 33, we find the roots and then calculate c as the negative product of the roots.
44 Medium
44 of 914
Application +4 | -1
If the sum of the squares of the roots of the equation x2−(a−2)x−(a+1)=0{x^2} - \left( {a - 2} \right)x - \left( {a + 1} \right) = 0 then the value of aa is
View answer and explanation
Correct option: D
1
Option D (1) is correct because the sum of the squares of the roots is minimized when a=1, as derived from completing the square.
45 Easy
45 of 914
Analyzing +4 | -1
If the roots of the equation bx2+cx+a=0b x^2+c x+a=0 be imaginary, then for all real values of xx, the expression 3b2x2+6bcx+2c23 b^2 x^2+6 b c x+2 c^2, is
View answer and explanation
Correct option: D
greater than (−4ab)(-4 a b)
Option D (greater than (−4ab)(-4 a b)) is correct because the minimum value of the expression is greater than −c2-c^2, and since the roots of the quadratic are imaginary, −c2>−4ab-c^2 > -4ab.

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46 Easy
46 of 914
Application +4 | -1
The number of real solutions of the equation x2−∣x∣−12=0{x^2} - |x| - 12 = 0 is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the equation simplifies to an absolute value equation, yielding two real solutions.
47 Medium
47 of 914
Analyzing +4 | -1
If the equation ax2+2  bx−3  c=0a{x^2} + 2\;bx - 3\;c = 0 has no real roots and 3  c4<a+b,\frac{{3\;c}}{4} < a + b, then
View answer and explanation
Correct option: A
c<0c < 0
Option A (c<0c < 0) is correct because the equation has no real roots, implying the quadratic is always positive or always negative, and given the condition, it must be always positive, leading to c<0c < 0.
48 Medium
48 of 914
Analyzing +4 | -1
If α\alpha and β\beta are the roots of 6x2−6x+1=06 x^2-6 x+1=0, then the value of 12[a+bα+cα2+dα3]+12[a+bβ+cβ2+dβ3]\frac{1}{2}\left[a+b \alpha+c \alpha^2+d \alpha^3\right]+\frac{1}{2}\left[a+b \beta+c \beta^2+d \beta^3\right] is:
View answer and explanation
Correct option: B
a1+b2+c3+d4\frac{a}{1}+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}
Option B (a1+b2+c3+d4\frac{a}{1}+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}) is correct because it is the simplified form of the given expression after substituting the values of the sum and product of the roots.
49 Medium
49 of 914
Analyzing +4 | -1
If x2−(a−3)x+a=0x^2-(a-3) x+a=0 has atleast one positive root, then
View answer and explanation
Correct option: C
a∈(−∞,0]∪[9,∞)a \in(-\infty, 0] \cup[9, \infty)
Option C (a∈(−∞,0]∪[9,∞)a \in(-\infty, 0] \cup[9, \infty)) is correct because it includes all values of 'a' for which the quadratic equation has at least one positive root, considering both cases where one or both roots are positive.
50 Medium
50 of 914
Analyzing +4 | -1
The solution set of the inequation 5(1/4)(log⁡5x)2  ≥5  x(1/5)(log⁡5x){5^{\left( {1/4} \right){{\left( {{{\log }_5}x} \right)}^2}\;}} \ge 5\;{x^{\left( {1/5} \right)\left( {{{\log }_5}x} \right)}}, is
View answer and explanation
Correct option: C
(0,5−25]∪[525,∞)\left( {0,{5^{ - 2\sqrt 5 }}\left] \cup \right[{5^{2\sqrt 5 }},\infty } \right)
Option A ((2−a)(2−b)(2−c)≥8abc\left( {2 - a} \right)\left( {2 - b} \right)\left( {2 - c} \right) \ge 8abc) is correct because it is derived from applying the AM-GM inequality to the given conditions.
51 Medium
51 of 914
Analyzing +4 | -1
Number of solutions for x2−2−2[x]=0x^2-2-2[x]=0, (where, [] denotes the greatest integer function) is
View answer and explanation
Correct option: B
1
Option B (1) is correct because only one solution, x=6x = \sqrt{6}, satisfies the given equation and the greatest integer function condition.
52 Easy
52 of 914
Application +4 | -1
Given that, for all x∈Rx \in R, the expression x2−2x+4x2+2x+4\frac{x^2-2 x+4}{x^2+2 x+4} lies between 13\frac{1}{3} and 3, the values between which the expression 9⋅32x+6⋅3x+49⋅32x−6⋅3x+4\frac{9 \cdot 3^{2 x}+6 \cdot 3^x+4}{9 \cdot 3^{2 x}-6 \cdot 3^x+4} lies, are
View answer and explanation
Correct option: B
32\frac{3}{2} and 22
The correct answer is B (3/2 and 2) because by substituting t=3x+1t = 3^{x+1} and using the given range for the expression, we can find the range for the new expression.
53 Easy
53 of 914
Application +4 | -1
If a,b,ca,b,c are all positive and in H.P., then the roots of ax2+2  bx+c=0a{x^2} + 2\;bx + c = 0 are
View answer and explanation
Correct option: B
Imaginary
Option B (Imaginary) is correct because the discriminant of the quadratic equation is negative, indicating imaginary roots.
54 Medium
54 of 914
Analyzing +4 | -1
If α\alpha and β\beta are the roots of the equation x2+px+q=0{x^2} + px + q = 0 and if the sum (α+β)x−α2+β22.x2+α3+β33.x3−...\left( {\alpha + \beta } \right)x - \frac{{{\alpha ^2} + {\beta ^2}}}{2}.{x^2} + \frac{{{\alpha ^3} + {\beta ^3}}}{3}.{x^3} - ... exists then it is equal to
View answer and explanation
Correct option: D
log⁡(1−px+qx2)\log \left( {1 - px + q{x^2}} \right)
Option D (log⁡(1−px+qx2)\log \left( {1 - px + q{x^2}} \right)) is correct because the given series can be expressed as the sum of two logarithmic series, which simplifies to the logarithm of a quadratic expression involving p, q, and x.
55 Easy
55 of 914
Analyzing +4 | -1
If the roots of the quadratic equation x−mmx+1=x+nnx+1\frac{x-m}{m x+1}=\frac{x+n}{n x+1} are reciprocal to each other, then:
View answer and explanation
Correct option: A
n=0n=0
Option A (n=0n=0) is correct because when the roots are reciprocal, their product is 1, leading to the equation m−n=m+nm-n = m+n, which simplifies to n=0n=0.
56 Easy
56 of 914
Application +4 | -1
Sum of the roots of the equation is 4(x−1x)2−4(x−1x)+1=04\left(x-\frac{1}{x}\right)^2-4\left(x-\frac{1}{x}\right)+1=0
View answer and explanation
Correct option: B
1
Option D (-1) is correct because the sum of the roots of the equation is 1.
57 Easy
57 of 914
Application +4 | -1
The roots of the equation ∣2x−1∣2−3∣2x−1∣+2=0{\left| {2x - 1} \right|^2} - 3\left| {2x - 1} \right| + 2 = 0 are
View answer and explanation
Correct option: D
{−12,0,1,32}\left\{ { - \frac{1}{2},0,1,\frac{3}{2}} \right\}
Option D ({−12,0,1,32}\left\{ { - \frac{{1}}{{2}},0,1,\frac{{3}}{{2}}} \right\}) is correct because these are the solutions obtained by solving the equation after substituting t=∣2x−1∣t = \left| {2x - 1} \right|.
58 Easy
58 of 914
Application +4 | -1
If α\alpha and β\beta be the roots of equation ax2+bx+c=0a x^2+b x+c=0, then lim⁡x→α(1+ax2+bx+c)1x−α\lim _{x \rightarrow \alpha}\left(1+a x^2+b x+c\right)^{\frac{1}{x-\alpha}} is equal to
View answer and explanation
Correct option: C
ea(α−β)e^{a(\alpha-\beta)}
Option C (ea(α−β)e^{a(\alpha-\beta)}) is correct because it represents the limit of the given expression as x approaches α, using L'Hopital's rule and the properties of roots of a quadratic equation.
59 Medium
59 of 914
Analyzing +4 | -1
The set of all a∈IRa \in I R for which the equation x∣x−1∣+∣x+2∣+a=0x|x-1|+|x+2|+a=0 has exactly one real root, is
View answer and explanation
Correct option: B
(−∞,∞)(-\infty, \infty)
The correct answer is B ((−∞,∞)(-\infty, \infty)) because for any value of 'a', the equation will always have exactly one real root due to the nature of the piecewise function.
60 Easy
60 of 914
Analyzing +4 | -1
For any real xx, the expression 2(k−x)[x+x2+k2]2(k-x)\left[x+\sqrt{x^2+k^2}\right] cannot exceed
View answer and explanation
Correct option: B
2k22 k^2
The correct answer is B (2k22 k^2) because it represents the maximum value that the expression can attain for any real x.
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