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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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871 Easy
871 of 914
Analyzing +4 | -1
Real roots of the equation k,  x2+5∣x∣+4=0k,\;{x^2} + 5\left| x \right| + 4 = 0 are
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the equation x2+5∣x∣+4=0{x^2} + 5\left| x \right| + 4 = 0 has no real roots since the absolute value terms are always non-negative, making the expression always positive.
872 Medium
872 of 914
Application +4 | -1
If 2α=−1−i32\alpha = - 1 - i\sqrt 3 and 2β=−1+i32\beta = - 1 + i\sqrt 3, then 5α4+5β4+7α−1β−15{\alpha ^4} + 5{\beta ^4} + 7{\alpha ^{ - 1}}{\beta ^{ - 1}} is equal to
View answer and explanation
Correct option: D
2
Option D (2) is correct because, after substituting the given values of α\alpha and β\beta and simplifying, the expression evaluates to 2.
873 Easy
873 of 914
Analyzing +4 | -1
The polynomial (ax2+bx+c)(ax2−dx−c),  ac≠0\left( {a{x^2} + bx + c} \right)\left( {a{x^2} - dx - c} \right),\;ac \ne 0 has
View answer and explanation
Correct option: B
At least two real roots
Option B (−2n1−α- \frac{{2n}}{{1 - \alpha }}) is correct because it is the result of manipulating the given series and using the properties of the nth root of unity.
874 Medium
874 of 914
Analyzing +4 | -1
If aa and bb are positive and unequal, then
View answer and explanation
Correct option: A
a4+b4>a3b+ab3a^4+b^4 > a^3 b+a b^3
Option A (a4+b4>a3b+ab3a^4+b^4 > a^3 b+a b^3) is correct because the difference between the two expressions can be factored into a positive term, proving the inequality.
875 Easy
875 of 914
Analyzing +4 | -1
The product of real roots of the equation x2+18x+30=2x2+18x+45,{x^2} + 18x + 30 = 2\sqrt {{x^2} + 18x + 45} , is
View answer and explanation
Correct option: B
20
Option B (20) is correct because after substitution and solving the quadratic equation, the product of the real roots is found to be 20.
876 Easy
876 of 914
Analyzing +4 | -1
Let SS be the set of all real roots of the equation, 3x(3x−1)+2=∣3x−1∣+∣3x−2∣.{3^x}({3^x} - 1) + 2 = \left| {{3^x} - 1} \right| + \left| {{3^x} - 2} \right|. Then SS
View answer and explanation
Correct option: A
is a singleton
Option A (is a singleton) is correct because after substituting t=3xt = 3^x, the resulting equation has only one real solution for tt, which leads to one real solution for xx.
877 Medium
877 of 914
Analyzing +4 | -1
If for x∈R,13<x2−2x+4x2+2x+4<3,x \in R,\frac{1}{3} < \frac{{{x^2} - 2x + 4}}{{{x^2} + 2x + 4}} < 3, then 9.32x−6.3x+49.32x+6.3x+4  \frac{{{{9.3}^{2x}} - {{6.3}^x} + 4}}{{{{9.3}^{2x}} + {{6.3}^x} + 4}}\; lies between
View answer and explanation
Correct option: B
13\frac{1}{3} and 3
Option B (13\frac{1}{3} and 3) is correct because the given expression can be transformed into the form of the inequality, and thus lies between the same bounds.
878 Medium
878 of 914
Application +4 | -1
Let f(x)=x2+2x+5f(x)=x^2+2 x+5 and α,β\alpha, \beta be roots of f(1t)=0f\left(\frac{1}{t}\right)=0, then α+β=\alpha+\beta=
View answer and explanation
Correct option: A
−25\frac{-2}{5}
Option A (-2/5) is correct because it is the sum of the roots of the transformed quadratic equation.
879 Easy
879 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of the equation 2x2−3x+4=02 x^2-3 x+4=0, then the equation whose roots are α2\alpha^2 and β2\beta^2 is:
View answer and explanation
Correct option: A
4x2+7x+16=04 x^2+7 x+16=0
Option A (4x2+7x+16=04 x^2+7 x+16=0) is correct because it is derived by finding the sum and product of the new roots (α2\alpha^2 and β2\beta^2) and constructing the quadratic equation.
880 Easy
880 of 914
Application +4 | -1
If α,β\alpha, \beta are the roots the quadratic equation x2−(3+2log⁡23−3log⁡32)x−2(3log⁡32−2log⁡23)x^2-\left(3+2^{\sqrt{\log _2 3}}-3 \sqrt{\log _3 2}\right) x-2\left(3^{\log _3 2}-2^{\log _2 3}\right) =0=0, then the value of α2+αβ+β2\alpha^2+\alpha \beta+\beta^2 is equal to
View answer and explanation
Correct option: A
7
Option A (7) is correct because simplifying the equation and finding the roots allows us to calculate the value of the expression alpha2 + alpha*beta + beta2.
881 Easy
881 of 914
Analyzing +4 | -1
If at least one root of 2x2+3x+5=0    and  ax2+bx+c=0,  a,  b,  c∈N2{x^2} + 3x + 5 = 0\;\,\,\mathrm{and\;}a{x^2} + bx + c = 0,\;a,\;b,\;c \in N is common, then the maximum value of a+b+ca + b + c is
View answer and explanation
Correct option: C
Does not exist
Option C (Does not exist) is correct because the roots of the first equation are imaginary, implying both roots must be common, leading to proportional coefficients, and thus no maximum value for a+b+c.
882 Easy
882 of 914
Analyzing +4 | -1
If a,b,ca,b,c are positive real numbers, then the number of real roots of the equation ax2+b∣x∣+c=0a{x^2} + b\left| x \right| + c = 0 is
View answer and explanation
Correct option: C
0
Option C (0) is correct because, with positive coefficients, the expression is always positive, thus never equals zero.
883 Easy
883 of 914
Application +4 | -1
If the roots of the given equation (cos⁡p−1)x2+(cos⁡p)x+sin⁡p=0\left( {\cos p - 1} \right){x^2} + \left( {\cos p} \right)x + \sin p = 0 are real, then
View answer and explanation
Correct option: C
p∈(0,  π)p \in \left( {0,\;\pi } \right)
Option C (p∈(0,π)p \in (0, \pi)) is correct because the discriminant of the quadratic must be non-negative for real roots, and this condition is satisfied within the given interval.
884 Medium
884 of 914
Analyzing +4 | -1
If pp and qq are odd prime numbers, then which of the following statements about the roots of the equation x2+2px−4q=0x^2+2 p x-4 q=0 is correct?
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Correct option: B
Always irrational
Option B (Always irrational) is correct because the discriminant is positive but not a perfect square, ensuring irrational roots.
885 Easy
885 of 914
Application +4 | -1
If a+b+c=0a + b + c = 0, then the roots of the equation 4ax2+3bx+2c=04a{x^2} + 3bx + 2c = 0 are
View answer and explanation
Correct option: C
Real
Option C (Real) is correct because the discriminant of the quadratic equation, after substituting b=−(a+c)b = -(a+c), simplifies to a form that is always non-negative, ensuring real roots.

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886 Easy
886 of 914
Application +4 | -1
If c>0c > 0 and 4a+c<2b4 a+c < 2 b, then ax2−bx+c=0a x^{2}-b x+c=0 has a root in the interval
View answer and explanation
Correct option: A
(0,2)(0,2)
Option A ((0,2)(0,2)) is correct because f(0)>0f(0) > 0 and f(2)<0f(2) < 0, indicating a root exists within the interval (0,2) by the Intermediate Value Theorem.
887 Easy
887 of 914
Application +4 | -1
If a,b∈{1,2,3,4}a, b \in\{1,2,3,4\} the number of quadratic equation of the form ax2+bx+c=0a x^2+b x+c=0 which have non-real complex roots is:
View answer and explanation
Correct option: C
52
Option C (52) is correct because we calculate the total number of possible quadratic equations and subtract the number of equations with real roots.
888 Easy
888 of 914
Analyzing +4 | -1
Let SS denote the set of all real values of aa for which the roots of the equation x2−2ax+a2−1=0{x^2} - 2ax + {a^2} - 1 = 0 lie between 5 and 10, then SS equals
View answer and explanation
Correct option: D
(6,  9)\left( {6,\;9} \right)
Option D ((6,  9)\left( {6,\;9} \right)) is correct because the roots of the equation are a+1 and a-1, and for these roots to lie between 5 and 10, 'a' must lie between 6 and 9.
889 Easy
889 of 914
Analyzing +4 | -1
The inequality 2x<3\frac{2}{x} < 3 is true, when xx belongs to
View answer and explanation
Correct option: C
(2/3,∞)∪(−∞,0)(2 / 3, \infty) \cup(-\infty, 0)
Option C ((2/3,∞)∪(−∞,0)2 / 3, \infty) \cup(-\infty, 0)) is correct because solving the inequality 2x<3\frac{2}{x} < 3 leads to this set of values for xx.
890 Easy
890 of 914
Application +4 | -1
If log⁡27(log⁡2x)=13{\log _{27}}\left( {{{\log }_2}x} \right) = \frac{1}{3}, then the value of xx is
View answer and explanation
Correct option: D
27
Option D (27) is correct because it is the value of x obtained by solving the given logarithmic equation.
891 Easy
891 of 914
Application +4 | -1
If 2+5i2+\sqrt{5} i is a root of x2−px+q=0x^2-p x+q=0 where pp and qq are real, then the ordered pair (p,q)(p, q) is equal to
View answer and explanation
Correct option: A
(4,9)(4,9)
Option A ((4,9)) is correct because since p and q are real, the other root is the conjugate, and we can find p and q using the sum and product of the roots.
892 Easy
892 of 914
Application +4 | -1
Rational roots of the equation 2x4+x3−11x2+x+2=02{x^4} + {x^3} - 11{x^2} + x + 2 = 0 are
View answer and explanation
Correct option: A
12\frac{1}{2} and 2
Option A (12\frac{1}{2} and 2) is correct because these are the only rational roots of the given quartic equation, found by reducing it to a quadratic.
893 Medium
893 of 914
Analyzing +4 | -1
If a,ba, b and cc are all distinct, then a(x−b)(x−c)(a−b)(a−c)+b(x−c)(x−a)(b−c)(b−a)+c(x−a)(x−b)(c−a)(c−b)−x is equal to a \frac{(x-b)(x-c)}{(a-b)(a-c)}+b \frac{(x-c)(x-a)}{(b-c)(b-a)}+c \frac{(x-a)(x-b)}{(c-a)(c-b)}-x \text { is equal to }
View answer and explanation
Correct option: A
0
Option A (0) is correct because the given expression is a quadratic polynomial that evaluates to zero at three distinct points, implying it must be identically zero.
894 Easy
894 of 914
Application +4 | -1
If x1x_1 and x2x_2 are the arithmetic and harmonic means of the roots of the equation ax2+bx+c=0a x^2+b x+c=0, the quadratic equation whose roots are x1x_1 and x2x_2, is
View answer and explanation
Correct option: B
2abx2+(b2+4ac)x+2bc=02 a b x^2+\left(b^2+4 a c\right) x+2 b c=0
Option B 2abx2+(b2+4ac)x+2bc=02 a b x^2+\left(b^2+4 a c\right) x+2 b c=0 is correct because it is the quadratic equation formed using the arithmetic and harmonic means of the roots of the given equation.
895 Easy
895 of 914
Analyzing +4 | -1
The value of xx in the given equation 4x−3x−12=3x+12−22x−1  {4^x} - {3^{x - \frac{1}{2}}} = {3^{x + \frac{1}{2}}} - {2^{2x - 1}}\; is
View answer and explanation
Correct option: B
32\frac{3}{2}
Option B (32\frac{3}{2}) is correct because by simplifying and taking logarithms, we find that x = 32\frac{3}{2} satisfies the given equation.
896 Medium
896 of 914
Application +4 | -1
The solution set of the inequation 2∣x−4∣>1,  x≠4,\frac{2}{{\left| {x - 4} \right|}} > 1,\;x \ne 4, is
View answer and explanation
Correct option: B
(2,  4)∪(4,  6)\left( {2,\;4} \right) \cup \left( {4,\;6} \right)
Option B ((2,  4)∪(4,  6)\left( {2,\;4} \right) \cup \left( {4,\;6} \right)) is correct because it represents the interval where the inequality holds true, excluding the point where the expression is undefined.
897 Easy
897 of 914
Understanding +4 | -1
The number of real roots of the equation, (x−2)2+(x−3)2+(x−4)2=0(x-2)^2+(x-3)^2+(x-4)^2=0 is -
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of squares can only be zero if each term is zero, which is impossible in this case.
898 Easy
898 of 914
Analyzing +4 | -1
If x=12(3+13)x = \frac{1}{2}\left( {\sqrt 3 + \frac{1}{{\sqrt 3 }}} \right), then x2−1x−x2−1\frac{{\sqrt {{x^2} - 1} }}{{x - \sqrt {{x^2} - 1} }} is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because by simplifying the expression for x and substituting it into the given expression, the result is 1.
899 Hard
899 of 914
Analyzing +4 | -1
If a,ba, b and c∈Rc \in R, then the square root of a2+b2+c2−bc−ca−aba^2+b^2+c^2-b c-c a-a b is greater than or equal to
View answer and explanation
Correct option: A
32max⁡{∣b−c∣,∣c−a∣,∣a−b∣}\frac{\sqrt{3}}{2} \max \{|b-c|,|c-a|,|a-b|\}
Option A (32max⁡∣b−c∣,∣c−a∣,∣a−b∣\frac{\sqrt{3}}{2} \max \\{|b-c|,|c-a|,|a-b|\\}) is correct because it is derived by manipulating the given expression and comparing it with the maximum difference between the variables.
900 Hard
900 of 914
Application +4 | -1
If α,  β\alpha ,\;\beta are the roots of the equation ax2+bx+c=0   such  that  β<α<0,a{x^2} + bx + c = 0\;\,\mathrm{such\;that}\;\beta < \alpha < 0, then the quadratic equation whose roots are ∣α∣,  ∣β∣,  \left| \alpha \right|,\;\left| \beta \right|,\; is given by
View answer and explanation
Correct option: B
ax2−∣b∣x+c=0a{x^2} - \left| b \right|x + c = 0
Option B (ax2−∣b∣x+c=0a{x^2} - \left| b \right|x + c = 0) is correct because it represents the quadratic equation with roots ∣α∣\left| \alpha \right| and ∣β∣\left| \beta \right| given that both α\alpha and β\beta are negative.
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