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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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211 Easy
211 of 914
Application +4 | -1
If 13−4i\frac{1}{3-4 i} is a root of ax2+bx+c=0,(a,ba x^2+b x+c=0,(a, b, c∈R,a≠0c \in \mathbf{R}, a \neq 0 ), then
View answer and explanation
Correct option: A
b+6c=0b+6 c=0
Option A (b+6c=0b+6c=0) is correct because, since a, b, and c are real, the complex conjugate of the root must also be a root, allowing us to find relationships between the coefficients.
212 Medium
212 of 914
Understanding +4 | -1
If p,q,rp,q,r are any real numbers, then
View answer and explanation
Correct option: B
min⁡(p,q)=12(P+q−∣p−q∣)\min \left( {p,q} \right) = \frac{1}{2}\left( {P + q - \left| {p - q} \right|} \right)
Option B is correct because it provides the correct formula for calculating the minimum of two real numbers p and q using the absolute value function.
213 Medium
213 of 914
Analyzing +4 | -1
Number of non-negative integral solution of equation y4+6xy2−8x=0y^4+6 x y^2-8 x=0 is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because only one non-negative integer value of y (y=0) results in a non-negative integer value for x.
214 Medium
214 of 914
Analyzing +4 | -1
If 0<a<1,0 < a < 1, then the solution set of the inequation 1+(log⁡ax)21+(log⁡ax)>1,\frac{{1 + {{\left( {{{\log }_a}x} \right)}^2}}}{{1 + \left( {{{\log }_a}x} \right)}} > 1, is
View answer and explanation
Correct option: C
(1,  1/a)∪(0,a)\left( {1,\;1/a} \right) \cup \left( {0,a} \right)
Option C ((1,  1/a)∪(0,a)\left( {1,\;1/a} \right) \cup \left( {0,a} \right)) is correct because solving the logarithmic inequality results in two intervals for x: (1, 1/a) and (0, a).
215 Easy
215 of 914
Application +4 | -1
If the difference between the roots of x2+ax−b=0{x^2} + ax - b = 0 is equal to the difference between the roots of x2−px+q=0{x^2} - px + q = 0, then a2−p2{a^2} - {p^2} in terms of bb and qq is
View answer and explanation
Correct option: A
−4(b+q)- 4\left( {b + q} \right)
Option A (x−axisx - axis) is correct because simplifying the given equation leads to y = 0, which represents the x-axis.
216 Medium
216 of 914
Application +4 | -1
If pp and qq are non-zero real numbers and α3+β3=−p,αβ=q\alpha^3+\beta^3=-p, \alpha \beta=q, then a quadratic equation whose roots are α2β,β2α\frac{\alpha^2}{\beta}, \frac{\beta^2}{\alpha} is :
View answer and explanation
Correct option: B
qx2+px+q2=0q x^{2}+p x+q^{2}=0
Option B (qx2+px+q2=0q x^{2}+p x+q^{2}=0) is correct because it's derived from finding the sum and product of the new roots and constructing the quadratic equation.
217 Easy
217 of 914
Analyzing +4 | -1
If a+b+c=0a + b + c = 0 and a≠ca \ne c then the roots of the equation (b+c−a)x2+(c+a−b)x+(a+b−c)=0,\left( {b + c - a} \right){x^2} + \left( {c + a - b} \right)x + \left( {a + b - c} \right) = 0, are
View answer and explanation
Correct option: A
Real and unequal
Option A (Real and unequal) is correct because the discriminant of the quadratic equation is positive, indicating two distinct real roots.
218 Medium
218 of 914
Application +4 | -1
Suppose a∈Ra \in \mathbf{R}. If 3x2+2(a2+1)x+(a23 x^2+2\left(a^2+1\right) x+\left(a^2\right. −3a+2)=0-3 a+2)=0 possesses roots of opposite signs, then aa lies in the interval:
View answer and explanation
Correct option: C
(1,2)(1,2)
Option C ((1,2)) is correct because the product of the roots must be negative for them to have opposite signs, leading to the inequality (a-1)(a-2) < 0, which is satisfied when a is between 1 and 2.
219 Easy
219 of 914
Application +4 | -1
If the roots of the equation (a2+b2)t2−2(ac+bd)t+(c2+d2)=0\left(a^2+b^2\right) t^2-2(a c+b d) t+\left(c^2+d^2\right)=0 are equal, then
View answer and explanation
Correct option: D
a/b=c/da / b=c / d
Option D (a/b=c/da/b = c/d) is correct because the condition for equal roots leads to the equation (ad−bc)2=0(ad-bc)^2 = 0, which simplifies to a/b=c/da/b = c/d.
220 Medium
220 of 914
Analyzing +4 | -1
If f(x)=x2+2bx+2c2  and  g(x)=−x2−2cx+b2  such  thatmin⁡f(x)>max⁡g(x),f\left( x \right) = {x^2} + 2bx + 2{c^2}\;\mathrm{and\;g}\left( x \right) = - {x^2} - 2cx + {b^2}\;\mathrm{such\;that}\min f\left( x \right) > \max \mathrm{g}\left( x \right), then the relation between bb and cc is
View answer and explanation
Correct option: D
∣c∣>∣b∣2\left| c \right| > \left| b \right|\sqrt 2
Option D (∣c∣>∣b∣2\left| c \right| > \left| b \right|\sqrt 2) is correct because the condition min⁡f(x)>max⁡g(x)\min f(x) > \max g(x) leads to the inequality ∣c∣>∣b∣2\left| c \right| > \left| b \right|\sqrt 2.
221 Easy
221 of 914
Application +4 | -1
If x2+2ax+10−3a>0{x^2} + 2ax + 10 - 3a > 0 for all x∈Rx \in R, then
View answer and explanation
Correct option: A
−5<a<2- 5 < a < 2
The correct answer is A (−5<a<2- 5 < a < 2) because for the quadratic to be always positive, the discriminant must be negative.
222 Medium
222 of 914
Analyzing +4 | -1
If the equation x2+5x+8=0x^2+5 x+8=0 and ax2+bx+c=0;a,b,c∈Ra x^2+b x+c=0 ; a, b, c \in R have a common root, then a:b:ca: b: c is
View answer and explanation
Correct option: B
1:5:81: 5: 8
Option B (1:5:81: 5: 8) is correct because the given quadratic equation has imaginary roots, implying the second equation shares the same roots, leading to proportional coefficients.
223 Easy
223 of 914
Application +4 | -1
The set of all integral values of xx for which 5x−1<(x+1)2<7x−3,5x - 1 < {\left( {x + 1} \right)^2} < 7x - 3, is
View answer and explanation
Correct option: D
{3}\left\{ 3 \right\}
Option D ({3}) is correct because solving the inequalities yields an interval for x, and the only integer within that interval is 3.
224 Easy
224 of 914
Application +4 | -1
The roots of the equation a(x2+1)−(a2+1)x=0a\left(x^2+1\right)-\left(a^2+1\right) x=0 are:
View answer and explanation
Correct option: A
a,1aa, \frac{1}{a}
Option A (a,1aa, \frac{1}{a}) is the correct answer because these values satisfy the given quadratic equation when substituted for x.
225 Medium
225 of 914
Analyzing +4 | -1
The equation x2⋅2∣x−3∣+4+2x−1x^2 \cdot 2^{|x-3|+4}+2^{x-1} =x2⋅2x+1+2∣x−3∣+2=x^2 \cdot 2^{x+1}+2^{|x-3|+2} has
View answer and explanation
Correct option: C
x=±12x= \pm \frac{1}{2} and x≥3x \geq 3
Option D (x=±12x= \pm \frac{1}{2} and x≥3x \geq 3) is correct because it represents the complete solution set obtained by considering both cases when x-3 is positive and negative.

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226 Medium
226 of 914
Application +4 | -1
The inequality ∣2x−3∣<1|2 x-3| < 1 is valid when xx lies in the interval
View answer and explanation
Correct option: B
(1,2)(1,2)
The correct answer is B ((1,2)) because solving the absolute value inequality ∣2x−3∣<1|2x-3| < 1 results in the interval (1, 2).
227 Medium
227 of 914
Application +4 | -1
If x2+2ax+10−3a>0{x^2} + 2ax + 10 - 3a > 0 for all x∈R,x \in R, then
View answer and explanation
Correct option: A
−5<a<2- 5 < a < 2
Option A (−5<a<2- 5 < a < 2) is correct because the given quadratic expression is always positive if its discriminant is negative, leading to this range for 'a'.
228 Easy
228 of 914
Application +4 | -1
If p1\mathrm{p}_1 and p2\mathrm{p}_2 are the lengths of the perpendiculars from the points (2,3,4)(2,3,4) and (1,1,4)(1,1,4) respectively on the plane 3x−6y+2z+11=03 x-6 y+2 z+11=0, then p1,p2p_1, p_2 are the roots of the equation
View answer and explanation
Correct option: B
7p2−23p+16=07 \mathrm{p}^2-23 \mathrm{p}+16=0
Option B (7p2−23p+16=07 p^2-23 p+16=0) is correct because it involves calculating the perpendicular distances from the given points to the plane and then forming a quadratic equation with those distances as roots.
229 Easy
229 of 914
Application +4 | -1
If the equation (a+1)x2−(a+2)x+(a+3)=0\left( {a + 1} \right){x^2} - \left( {a + 2} \right)x + \left( {a + 3} \right) = 0 has roots equal in magnitude but opposite in sign, then the roots of the equation are
View answer and explanation
Correct option: B
±12a\pm \frac{1}{2}a
Option B (±12a\pm \frac{1}{2}a) is correct because it satisfies the condition that the roots are equal in magnitude but opposite in sign, and the value of 'a' obtained from the coefficient of x.
230 Medium
230 of 914
Application +4 | -1
If a2+b2+c2=1,{a^2} + {b^2} + {c^2} = 1, then ab+bc+ca  ab + bc + ca\; lies in the interval
View answer and explanation
Correct option: A
[−12,1]\left[ { - \frac{1}{2},1} \right]
Option A [−12,1]\left[ { - \frac{1}{2},1} \right] is correct because it represents the interval in which ab+bc+caab + bc + ca lies, derived from the given condition and algebraic manipulations.
231 Easy
231 of 914
Application +4 | -1
A value of kk for which the quadratic equation x2−2x(l+3k)+7(2k+3)=0{x^2} - 2x\left( {\mathrm{l} + 3k} \right) + 7\left( {2k + 3} \right) = 0 has equal roots, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because when k=2, the discriminant of the quadratic equation is zero, indicating equal roots.
232 Medium
232 of 914
Analyzing +4 | -1
If a>0,a≠1a > 0, a \neq 1, then the equation 2log⁡xa+log⁡axa+3log⁡a2xa=02 \log _x a+\log _{a x} a+3 \log _{a^2 x} a=0 has
View answer and explanation
Correct option: B
two real roots
Option B (two real roots) is correct because solving the logarithmic equation leads to two distinct real solutions for x, given the constraints on a.
233 Easy
233 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of the equation 2x(2x+1)=1,2x\left( {2x + 1} \right) = 1, then β\beta is equal to
View answer and explanation
Correct option: B
−2α(α+1)- 2\alpha \left( {\alpha + 1} \right)
Option B (−2α(α+1)- 2\alpha (\alpha + 1)) is correct because it correctly relates the roots of the given quadratic equation.
234 Medium
234 of 914
Analyzing +4 | -1
If the equation x3−3x+a=0{x^3} - 3x + a = 0 has distinct roots between 0 and 1, then the value of aa is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the derivative of the given function does not have any roots between 0 and 1, implying that the function cannot have distinct roots between 0 and 1.
235 Medium
235 of 914
Analyzing +4 | -1
If α,β\alpha ,\beta are the roots of the equation λ(x2−x)+x+5=0\lambda \left( {{x^2} - x} \right) + x + 5 = 0 and if λ1  and  λ2{\lambda _1}\;\mathrm{and}\;{\lambda _2} are two values of λ\lambda obtained from αβ+βα=45\frac{\alpha }{\beta } + \frac{\beta }{\alpha } = \frac{4}{5} then λ1λ22+λ2λ12\frac{{{\lambda _1}}}{{\lambda _2^2}} + \frac{{{\lambda _2}}}{{\lambda _1^2}} is equal to
View answer and explanation
Correct option: D
4048
Option D (4048) is correct because it is the result of substituting the values of λ1+λ2\lambda_1 + \lambda_2 and λ1λ2\lambda_1 \lambda_2 into the expression after simplifying the given equation.
236 Easy
236 of 914
Application +4 | -1
If the roots of the equation 9x2+4ax+4=09 x^2+4 a x+4=0 are imaginary, then
View answer and explanation
Correct option: A
a∈(−3,3)a \in(-3,3)
Option A (a∈(−3,3)a \in(-3,3)) is correct because the roots are imaginary when the discriminant is less than zero, leading to the inequality −3<a<3-3 < a < 3.
237 Easy
237 of 914
Analyzing +4 | -1
The number of real roots of the equation (x−1)2+(x−2)2+(x−3)2=0{\left( {x - 1} \right)^2} + {\left( {x - 2} \right)^2} + {\left( {x - 3} \right)^2} = 0 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the sum of squares cannot be zero for real numbers unless each term is zero, which is not possible here.
238 Easy
238 of 914
Application +4 | -1
The real roots of the equation x2/3−x1/3−2=0{x^{2/3}} - {x^{1/3}} - 2 = 0 are
View answer and explanation
Correct option: D
1,−81, - 8
Option D (1,−81, - 8) is correct because by substituting y=x1/3y = x^{1/3}, we get a quadratic equation whose solutions lead to x=1x = 1 and x=−8x = -8.
239 Medium
239 of 914
Application +4 | -1
Let α,β\alpha, \beta be the roots of the equation x2−2x+2=0x^{2}-\sqrt{2} x+2=0. Then α14+β14\alpha^{14}+\beta^{14} is equal to
View answer and explanation
Correct option: C
−128-128
Option C (-128) is correct because after finding the roots in polar form and raising them to the power of 14, the sum simplifies to -128.
240 Medium
240 of 914
Application +4 | -1
The solution of the equation  (3+22)x2−8+(3+22)8−x2=6\,{\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} + {\left( {3 + 2\sqrt 2 } \right)^{8 - {x^2}}} = 6 are
View answer and explanation
Correct option: D
±3,  ±7\pm 3,\; \pm \sqrt 7
Option D (±3,;±7\pm 3,\\; \pm \sqrt 7) is correct because by substituting y=(3+22)x2−8y = {\left( {3 + 2\sqrt 2 } \right)^{{x^2} - 8}} and solving the quadratic equation, we find the possible values of x.
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