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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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Showing 841-870 of 914 questions

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841 Easy
841 of 914
Analyzing +4 | -1
The sum of non-real roots of the equation (x2+x−2)(x2+x−3)=12,\left( {{x^2} + x - 2} \right)\left( {{x^2} + x - 3} \right) = 12, is
View answer and explanation
Correct option: B
−1- 1
Option B (-1) is correct because the sum of the non-real roots of the given equation is -1, obtained by solving the equation and identifying the non-real roots.
842 Medium
842 of 914
Application +4 | -1
Let α, β\alpha ,\,\beta be two roots of the equation x2+(20)1/4x+(5)1/2=0{x^2} + {\left( {20} \right)^{1/4}}x + {\left( 5 \right)^{1/2}} = 0. Then, α8+β8{\alpha ^8} + {\beta ^8} is equal to
View answer and explanation
Correct option: C
50
Option C (50) is correct because by using Vieta's formulas and algebraic manipulations, we can find the value of alpha8 + beta8.
843 Easy
843 of 914
Analyzing +4 | -1
The equation x34(log⁡2x)2+(log⁡2x)−54=2{x^{\frac{3}{4}{{({{\log }_2}x)}^2} + {{\left( {{{\log }_2}x} \right)}^ - }\frac{5}{4}}} = \sqrt 2 has
View answer and explanation
Correct option: C
Exactly one irrational solution
Option C (Exactly one irrational solution) is correct because after solving the equation, only one of the three solutions is irrational.
844 Easy
844 of 914
Application +4 | -1
Solution of ∣3x+2∣<1|3 x+2| < 1 is
View answer and explanation
Correct option: C
(−1,−1/3)(-1,-1 / 3)
Option C ((−1,−1/3))((-1,-1 / 3)) is correct because it represents the interval of xx values that satisfy the absolute value inequality.
845 Easy
845 of 914
Application +4 | -1
If the difference of the roots of the equation x2−bx+c=0{x^2} - bx + c = 0 be 1, then
View answer and explanation
Correct option: A
b2−4c−1=0{b^2} - 4c - 1 = 0
Option A (b2−4c−1=0{b^2} - 4c - 1 = 0) is correct because using the difference of roots formula and Vieta's formulas leads to the given relationship.
846 Easy
846 of 914
Analyzing +4 | -1
If the equations ax2+bx+c=0a{x^2} + bx + c = 0 and x3+3  x2+3  x+2=0{x^3} + 3\;{x^2} + 3\;x + 2 = 0 have to common roots, then
View answer and explanation
Correct option: C
a=b=ca = b = c
Option C (a=b=ca = b = c) is correct because if the two equations have two common roots, then the coefficients must be proportional, leading to a=b=ca=b=c.
847 Hard
847 of 914
Application +4 | -1
If ai>0  {a_i} > 0\; for   i=1,2,…,n  \;i = 1,2, \ldots ,n\; and   a1a2…an=1,  \;{a_1}{a_2} \ldots {a_n} = 1,\; then minimum value of (1+a1)(1+a2)…(1+an)  \left( {1 + {a_1}} \right)\left( {1 + {a_2}} \right) \ldots \left( {1 + {a_n}} \right)\; is
View answer and explanation
Correct option: B
2n{2^n}
Option B (2n{2^n}) is correct because the minimum value is obtained by repeatedly applying AM ≥ GM inequality to each factor and using the given condition a1a2…an=1{a_1}{a_2} \ldots {a_n} = 1.
848 Medium
848 of 914
Application +4 | -1
If the equation 2x2+3x+5λ=0   and   x2+2x+3λ=02{x^2} + 3x + 5\lambda = 0\;\,\mathrm{and}\,\;{x^2} + 2x + 3\lambda = 0 have a common root, then λ\lambda is equal to
View answer and explanation
Correct option: C
0,  −10,\; - 1
The correct answer is C (0,  −10,\; - 1) because these are the values of λ\lambda for which the given quadratic equations have a common root.
849 Easy
849 of 914
Application +4 | -1
If one of the root of the equation x2+ax+3=0{x^2} + ax + 3 = 0 is 3 and one of the roots of the equation x2+ax+b=0{x^2} + ax + b = 0 is three times the other root, then the value of bb is equal to
View answer and explanation
Correct option: A
3
The correct answer is A (3) because solving for the roots and using the given condition leads to b = 3.
850 Medium
850 of 914
Analyzing +4 | -1
If a∈Ra \in R and the equation (a−2)(x−[x])2+2(x−[x])+a2=0(a-2)(x-[x])^2+2(x-[x])+a^2=0 (where, [x][x] denotes the greatest integer ≤x\leq x ) has no integral solution and has exactly one solution in (2,3)(2,3), then aa lies in the interval
View answer and explanation
Correct option: C
(−1,0)(-1,0)
Option C ((-1,0)) is correct because the given conditions lead to the inequality a(a+1)<0a(a+1) < 0, which is satisfied when aa lies in the interval (−1,0)(-1, 0).
851 Easy
851 of 914
Application +4 | -1
If 2+i32 + i\sqrt 3 is a root of the equation x2+px+q=0{x^2} + px + q = 0, then the value of (p,  q)\left( {p,\;q} \right) is
View answer and explanation
Correct option: B
(−4,7)\left( { - 4,\mathrm{ }7} \right)
Option B ((-4, 7)) is correct because the roots of the quadratic equation are complex conjugates, and the coefficients p and q can be found using Vieta's formulas.
852 Medium
852 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of 4x2+3x+7=04 x^2+3 x+7=0, then the value of 1α3+1β3\frac{1}{\alpha^3}+\frac{1}{\beta^3} is
View answer and explanation
Correct option: C
225/343225 / 343
Option C (225/343) is correct because it is the result of substituting the sum and product of roots into the simplified expression.
853 Medium
853 of 914
Analyzing +4 | -1
Let P(x)=0P(x)=0 be the polynomial equation of least possible degree with rational coefficients having 73+493\sqrt[3]{7}+\sqrt[3]{49} as a root. Then, the product of all the roots of P(x)=0P(x)=0 is
View answer and explanation
Correct option: A
56
Option A (56) is correct because the product of the roots of the derived cubic equation is equal to the constant term, which is 56.
854 Easy
854 of 914
Application +4 | -1
If (x−2)\left( {x - 2} \right) is a common factor of the expressions x2+ax+b    and   x2+cx+d,{x^2} + ax + b\;\,\,\mathrm{and}\,\;{x^2} + cx + d, then b−dc−a\frac{{b - d}}{{c - a}} is equal to
View answer and explanation
Correct option: D
2
Option D (2) is correct because substituting x=2 into both quadratic expressions and solving for the ratio (b-d)/(c-a) yields 2.
855 Easy
855 of 914
Application +4 | -1
If ∣2x−3∣=5|2 x-3|=5 then x=x=
View answer and explanation
Correct option: A
4,−14,-1
Option A (4,-1) is correct because by solving the absolute value equation, we find the two possible values of x to be 4 and -1.

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856 Easy
856 of 914
Application +4 | -1
In a quadratic equation with leading coefficient 1 , a student reads the coefficient 16 of xx wrongly as 19 and obtain the roots as -15 and -4 . The correct roots are
View answer and explanation
Correct option: B
−6,−10-6,-10
Option B (−6,−10-6,-10) is correct because the student incorrectly read the coefficient of x, but the constant term was read correctly, allowing us to determine the correct roots.
857 Hard
857 of 914
Analyzing +4 | -1
If p(x)=ax2+bx+c   and  Q(x)=−ax2+dx+cp\left( x \right) = a{x^2} + bx + c\,\;\mathrm{and}\;Q\left( x \right) = - a{x^2} + dx + c, where ac≠0ac \ne 0, then P(x)Q(x)=0P\left( x \right)Q\left( x \right) = 0 has at least
View answer and explanation
Correct option: B
Two real roots
Option B (Two real roots) is correct because if both quadratics had only imaginary roots, it would lead to a contradiction, thus at least one must have real roots.
858 Medium
858 of 914
Analyzing +4 | -1
If roots of the equation x4−8x3+bx2+cx+16=0x^4-8 x^3+b x^2+c x+16=0 are positive, then
View answer and explanation
Correct option: C
b=24,c=−32b=24, c=-32
Option D (b=24,c=−32b=24, c=-32) is correct because by using AM ≥ GM, all roots are found to be equal, allowing the determination of b and c by expanding (x−2)4(x-2)^4.
859 Easy
859 of 914
Application +4 | -1
The harmonic mean of the roots of the equation (5+2)x2−(4+5)x+8+25=0\left( {5 + \sqrt 2 } \right){x^2} - \left( {4 + \sqrt 5 } \right)x + 8 + 2\sqrt 5 = 0 is
View answer and explanation
Correct option: B
4
The correct answer is C ((1, 2)) because the roots have opposite signs when the product of the roots is negative, leading to k in the interval (1, 2).
860 Medium
860 of 914
Analyzing +4 | -1
The equations ax2+bx+a=0a{x^2} + bx + a = 0 and x3−2  x2+2  x−1=0{x^3} - 2\;{x^2} + 2\;x - 1 = 0 have 2 roots in common. Then, a+ba + b must be equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because by finding the roots of the cubic equation and substituting them into the quadratic equation, we can derive that a + b = 0.
861 Medium
861 of 914
Application +4 | -1
If for all real values of aa one root of the equation x2−3ax+f(a)=0x^2-3 a x+f(a)=0 is double of the other, f(x)f(x) is equal
View answer and explanation
Correct option: C
2x22 x^2
Option C 2x22 x^2 is correct because it represents the function f(x)f(x) derived from the given condition that one root is double the other for all real values of aa.
862 Medium
862 of 914
Application +4 | -1
If α\alpha is a root of x2+x+1=0x^2+x+1=0 satisfying (1+α)7=a+bα+cα2(1+\alpha)^7=a+b \alpha+c \alpha^2, then the ordered triplet (a,b,c)(a, b, c) is
View answer and explanation
Correct option: C
(3,3,2)(3,3,2)
Option C ((3,3,2)) is correct because it correctly identifies the ordered triplet by using the properties of complex cube roots of unity and substituting into the given equation.
863 Medium
863 of 914
Application +4 | -1
The roots of the given equation (p−q)x2+(q−r)x+(r−p)=0\left( {p - q} \right){x^2} + \left( {q - r} \right)x + \left( {r - p} \right) = 0 are
View answer and explanation
Correct option: C
r−pp−q,1\frac{{r - p}}{{p - q}},1
The correct answer is C (r−pp−q,1\frac{{r - p}}{{p - q}},1) because by inspection, x=1 is a root, and the other root can be found using the relationship between roots and coefficients.
864 Medium
864 of 914
Analyzing +4 | -1
The solution set of ∣x+3∣+xx+2>1\frac{|x+3|+x}{x+2} > 1 is
View answer and explanation
Correct option: C
(−5,−2)∪(−1,∞)(-5,-2) \cup(-1, \infty)
Option C ((−5,−2)∪(−1,∞)(-5,-2) \cup (-1, \infty)) is correct because it represents the solution set obtained by solving the inequality considering both cases of the absolute value.
865 Easy
865 of 914
Application +4 | -1
If roots of 7x2−11x+k=0,k≠07 x^2-11 x+k=0, k \neq 0 are reciprocal of each other, then kk is equal to
View answer and explanation
Correct option: C
7
Option C (7) is correct because if the roots are reciprocal, their product is 1, which equals c/a, thus k/7 = 1, so k = 7.
866 Easy
866 of 914
Application +4 | -1
Two non-integer roots of (x2−5x)2−7(x2−5x)+6=0\left(x^2-5 x\right)^2-7\left(x^2-5 x\right)+6=0 are
View answer and explanation
Correct option: A
12(5+29),12(5−29)\frac{1}{2}(5+\sqrt{29}), \frac{1}{2}(5-\sqrt{29})
Option A (12(5+29),12(5−29)\frac{1}{2}(5+\sqrt{29}), \frac{1}{2}(5-\sqrt{29})) is correct because these are the non-integer roots obtained by solving the quadratic equation after substitution.
867 Medium
867 of 914
Application +4 | -1
If α\alpha and β\beta are roots of the equation, x2−42kx+2e4ln⁡k−1=0x^{2}-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0 for some kk, and α2+β2=66\alpha^{2}+\beta^{2}=66, then α3+β3\alpha^{3}+\beta^{3} is equal to:
View answer and explanation
Correct option: D
−2802-280 \sqrt{2}
Option D (−2802-280 \sqrt{2}) is correct because solving for k and substituting it into the expression for α3+β3\alpha^{3}+\beta^{3} yields −2802-280 \sqrt{2}.
868 Easy
868 of 914
Analyzing +4 | -1
The solution of the equation 1+a+a2+a3+……+ax1+a+a^2+a^3+\ldots \ldots+a^x =(1+a)(1+a2)(1+a4)=(1+a)\left(1+a^2\right)\left(1+a^4\right) is given by xx is equal to:
View answer and explanation
Correct option: C
7
Option C (7) is correct because solving the given equation by simplifying the right-hand side and comparing exponents yields x=7.
869 Easy
869 of 914
Application +4 | -1
The solution set of the inequation x2+(a+b)x+ab<0,a<b,{x^2} + \left( {a + b} \right)x + ab < 0,a < b, is
View answer and explanation
Correct option: C
(−b,  −a)\left( { - b,\; - a} \right)
Option C ((−b,  −a)(- b,\; - a)) is correct because the solution to the inequality (x+a)(x+b)<0(x + a)(x + b) < 0 is the interval between the roots −b-b and −a-a, given that a<ba < b.
870 Easy
870 of 914
Application +4 | -1
Suppose aa is an integer and x1x_1 and x2x_2 are positive real roots of x2+(2a−1)x+a2=0x^2+(2 a-1) x+a^2=0, then value of ∣x1−x2∣\left|\sqrt{x_1}-\sqrt{x_2}\right| is
View answer and explanation
Correct option: A
1
Option A (1) is correct because simplifying the expression ∣x1−x2∣\left| \sqrt{{x_1}} - \sqrt{{x_2}} \right| using the given conditions results in a value of 1.
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