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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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811 Easy
811 of 914
Application +4 | -1
If α,β\alpha, \beta are the roots of the quadratic equation ax2+bx+c=0a x^2+b x+c=0, then the quadratic equation whose roots are α3,β3\alpha^3, \beta^3 is
View answer and explanation
Correct option: A
a3y2+(b3−3abc)y+c3=0a^3 y^2+\left(b^3-3 a b c\right) y+c^3=0
Option A (a3y2+(b3−3abc)y+c3=0a^3 y^2+\left(b^3-3 a b c\right) y+c^3=0) is correct because it is the transformed quadratic equation with roots α3\alpha^3 and β3\beta^3.
812 Medium
812 of 914
Analyzing +4 | -1
If α,  β\alpha ,\;\beta be the roots of x2−a(x−1)+b=0{x^2} - a\left( {x - 1} \right) + b = 0, then the value of 1α2−aα+1β2+aβ+2a+b\frac{1}{{{\alpha ^2} - a\alpha }} + \frac{1}{{{\beta ^2} + a\beta }} + \frac{2}{{a + b}} is
View answer and explanation
Correct option: C
0
Option C (0) is correct because by using the properties of roots and simplifying the expression, the value becomes zero.
813 Medium
813 of 914
Analyzing +4 | -1
If the equation ax2+2bx+3c=0   and  3x2+8x+15=0a{x^2} + 2bx + 3c = 0\;\,\mathrm{and}\;3{x^2} + 8x + 15 = 0 have a common root, where a,  b,  ca,\;b,\;c are the lengths of the sides of a ΔABC\Delta ABC, then sin⁡2A+sin⁡2B+sin⁡2C{\sin ^2}A + {\sin ^2}B + {\sin ^2}C is equal to
View answer and explanation
Correct option: D
2
Option B (x−c)(x−d)+k=0\left( {x - c} \right)\left( {x - d} \right) + k = 0 is correct because if c and d are roots of (x-a)(x-b) - k = 0, then (x-a)(x-b) - k = (x-c)(x-d). Thus, (x-c)(x-d) + k = (x-a)(x-b), implying a and b are roots of (x-c)(x-d) + k = 0.
814 Medium
814 of 914
Analyzing +4 | -1
If the difference between the corresponding roots of x2+x^2+ ax+b=0a x+b=0 and x2+bx+a=0x^2+b x+a=0 is same and a≠ba \neq b, then:
View answer and explanation
Correct option: A
a+b+4=0a+b+4=0
Option A (a+b+4=0a+b+4=0) is correct because it is the result of equating the difference between the roots of the two quadratic equations and simplifying the expression.
815 Medium
815 of 914
Analyzing +4 | -1
Let S={x:x∈R and (3+2)x2−4+(3−2)x2−4=10}S=\left\{\begin{array}{c}x: x \in \mathbb{R} \text { and }(\sqrt{3}+\sqrt{2})^{x^{2}-4} +(\sqrt{3}-\sqrt{2})^{x^{2}-4}=10\end{array}\right\}. Then n(S)n(S) is equal to
View answer and explanation
Correct option: D
6
Option B (4) is correct because solving the equation for x yields four distinct real solutions.
816 Easy
816 of 914
Application +4 | -1
If x2r2−r−6+y2r2−6r+5=1\frac{x^2}{r^2-r-6}+\frac{y^2}{r^2-6 r+5}=1 will represents the ellipse, if rr lies in the interval
View answer and explanation
Correct option: A
(−∞,−2)∪(5,∞)(-\infty,-2) \cup(5, \infty)
The correct answer is A ((−∞,−2)∪(5,∞))((-\infty,-2) \cup(5, \infty)). For the equation to represent an ellipse, both denominators must be positive.
817 Easy
817 of 914
Application +4 | -1
If α,β\alpha ,\beta are the roots of ax2+bx+c=0a{x^2} + bx + c = 0 the equation whose roots are 2+α,2+β2 + \alpha ,2 + \beta is
View answer and explanation
Correct option: D
ax2+x(b−4  a)+4  a−2  b+c=0a{x^2} + x\left( {b - 4\;a} \right) + 4\;a - 2\;b + c = 0
Option D (ax2+x(b−4  a)+4  a−2  b+c=0a{x^2} + x\left( {b - 4\;a} \right) + 4\;a - 2\;b + c = 0) is correct because by transforming the original equation using the relationship between the roots, we obtain the new equation with roots 2 + α\alpha and 2 + β\beta.
818 Easy
818 of 914
Application +4 | -1
Sum of the roots of the equation is

4x−3(2x+3)+128=04^x-3\left(2^{x+3}\right)+128=0 (1)
is
View answer and explanation
Correct option: C
7
Option C (7) is correct because the sum of the roots is obtained by substituting y=2xy = 2^x and solving the resulting quadratic equation for x.
819 Easy
819 of 914
Application +4 | -1
If aa is the minimum root of the equation x2−3∣x∣−2=0x^2-3|x|-2=0, then value of −1/a-1 / a is
View answer and explanation
Correct option: A
(17−3)/4(\sqrt{17}-3) / 4
Option A ((17−3)/4(\sqrt{{17}}-3) / 4) is correct because it is the negative reciprocal of the minimum root of the given equation after solving for |x| and considering both positive and negative roots.
820 Easy
820 of 914
Application +4 | -1
If the product 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 2, then the sum of roots is
View answer and explanation
Correct option: B
−1- 1
Option B (-1) is correct because by finding 'a' using the product of roots condition and substituting it into the sum of roots formula, we get -1.
821 Easy
821 of 914
Application +4 | -1
The values of aa and bb so that x4+12x3+46x2+ax+bx^{4}+12 x^{3}+46 x^{2}+a x+b is square of quadratic expression are, respectively,
View answer and explanation
Correct option: A
60,25
Option A (60, 25) is correct because when a=60a = 60 and b=25b = 25, the given expression becomes a perfect square of a quadratic expression.
822 Easy
822 of 914
Analyzing +4 | -1
If x∈R,x \in R, then the expression 9x−3x+1{9^x} - {3^x} + 1 assumes
View answer and explanation
Correct option: C
All real values greater than 34\frac{3}{4}
Option C (All real values greater than 34\frac{3}{4}) is correct because completing the square shows the minimum value of the expression is 34\frac{3}{4}.
823 Easy
823 of 914
Analyzing +4 | -1
The value of pp for which both the roots of the equation 4  x2−20  px+(25  p2+15  p−66)=04\;{x^2} - 20\;px + \left( {25\;{p^2} + 15\;p - 66} \right) = 0 are less than 2, lies in
View answer and explanation
Correct option: D
(−∞,−1)\left( { - \infty , - 1} \right)
Option D ((−∞,−1)\left( { - \infty , - 1} \right)) is correct because it satisfies the conditions for both roots to be less than 2.
824 Medium
824 of 914
Analyzing +4 | -1
For the equation ∣x∣2+∣x∣−6=0{\left| x \right|^2} + \left| x \right| - 6 = 0
View answer and explanation
Correct option: B
There are only two distinct roots
Option B (There are only two distinct roots) is correct because solving the equation results in two possible values for x: 2 and -2.
825 Medium
825 of 914
Understanding +4 | -1
If α\alpha and β(α<β)\beta (\alpha < \beta ) are two distinct roots of the equation ax2+bx+c=0,a{x^2} + bx + c = 0, then
View answer and explanation
Correct option: C
α<−b2a<β\alpha < - \frac{b}{{2a}} < \beta
Option C (α<−b2a<β\alpha < -\frac{b}{{2a}} < \beta) is correct because, by Rolle's theorem, there exists a root of the derivative between any two roots of the original equation, and the root of the derivative is -b/2a.

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826 Medium
826 of 914
Understanding +4 | -1
For positive real number a,b,c  a,b,c\; which one of the following holds?
View answer and explanation
Correct option: A
a2+b2+c2≥bc+ca+ab{a^2} + {b^2} + {c^2} \ge bc + ca + ab
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)) is correct because it correctly defines the minimum of two numbers using the absolute value function.
827 Easy
827 of 914
Application +4 | -1
If the equations ax2+bc+c=0a{x^2} + bc + c = 0 and 2x2+3x+4=02{x^2} + 3x + 4 = 0 have a common root, then a  :b  :ca\;:b\;:c
View answer and explanation
Correct option: A
2  :3  :42\;:3\;:4
The correct answer is A (2  :  3  :  42\;:\;3\;:\;4) because if two quadratic equations have a common root and one has complex roots, they must have both roots in common, implying their coefficients are proportional.
828 Easy
828 of 914
Analyzing +4 | -1
If the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0 of the form k+1k   and   k+2k+1,   then   (a+b+c)2\frac{{k + 1}}{k}\;\,\mathrm{and}\,\;\frac{{k + 2}}{{k + 1}},\;\,\mathrm{then}\,\;{\left( {a + b + c} \right)^2} is equal to
View answer and explanation
Correct option: C
b2−4ac{b^2} - 4ac
The correct answer is C (b2−4ac{b^2} - 4ac) because by using the given roots and Vieta's formulas, we can derive the relationship and simplify to find the value of (a+b+c)2(a+b+c)^2.
829 Easy
829 of 914
Analyzing +4 | -1
If α,β\alpha ,\beta are ther roots of ax2+bx+c=0a{x^2} + bx + c = 0 and α+β,α2+β2,α3+β3\alpha + \beta ,{\alpha ^2} + {\beta ^2},{\alpha ^3} + {\beta ^3} are in G.P., where Δ=−b2−4ac\Delta = - {b^2} - 4ac, then
View answer and explanation
Correct option: D
cΔ=0c\Delta = 0
Option D (cΔ=0c\Delta = 0) is correct because the given condition that α+β,α2+β2,α3+β3\alpha + \beta, \alpha^2 + \beta^2, \alpha^3 + \beta^3 are in G.P. leads to the equation ac(b2−4ac)=0ac(b^2 - 4ac) = 0, which simplifies to cΔ=0c\Delta = 0.
830 Medium
830 of 914
Analyzing +4 | -1
The value of aa for which the equations x3+ax+1=0   and  x4+ax2+1=0{x^3} + ax + 1 = 0\;\,\mathrm{and\;}{x^4} + a{x^2} + 1 = 0 have a common root, is
View answer and explanation
Correct option: A
−2- 2
Option A (-2) is correct because substituting x=1 (the common root) into the equation yields a = -2.
831 Easy
831 of 914
Application +4 | -1
The graph of the function y=16x2+8(a+5)x−7  a−5y = 16{x^2} + 8\left( {a + 5} \right)x - 7\;a - 5 is strictly above the xx-axis, then 'a' must satisfy the inequality
View answer and explanation
Correct option: A
−15<a<−2- 15 < a < - 2
Option A (-15 < a < -2) is correct because for the quadratic to be strictly above the x-axis, its discriminant must be negative, leading to the specified inequality for 'a'.
832 Easy
832 of 914
Application +4 | -1
If (1−p)\left( {1 - p} \right) is a root of quadratic equation x2+px+(1−p)=0{x^2} + px + \left( {1 - p} \right) = 0, then its roots are
View answer and explanation
Correct option: C
0,−10, - 1
Option C (0,−10, - 1) is correct because substituting (1−p)(1-p) into the quadratic equation and solving for pp, we find the roots of the equation.
833 Easy
833 of 914
Application +4 | -1
If a+b+c=0a+b+c=0 and a,b,ca, b, c are rational, then the roots of the equation
(b+c−a)x2+(c+a−b)x+(a+b−c)=0(b+c-a) x^2+(c+a-b) x+(a+b-c)=0 are
View answer and explanation
Correct option: A
Rational
Option A (Rational) is correct because the discriminant of the quadratic equation is a perfect square, implying rational roots.
834 Medium
834 of 914
Analyzing +4 | -1
For the equation 3x2+px+3=0,p>03 x^2+p x+3=0, p>0, if one of the roots is the square of the other, then the integral value of p=p=
View answer and explanation
Correct option: B
3
Option B (3) is correct because the condition that one root is the square of the other leads to a relationship between the roots and the coefficients, allowing us to solve for p.
835 Easy
835 of 914
Application +4 | -1
The set of all possible values of xx for which 13 is the A.M. of 51+x{5^{1 + x}} and 51−x,{5^{1 - x}}, is
View answer and explanation
Correct option: B
{−1,1}\left\{ { - 1,1} \right\}
Option B ({-1, 1}) is correct because these are the only values of x that satisfy the given arithmetic mean condition.
836 Easy
836 of 914
Analyzing +4 | -1
The equation ex−x−1=0e^x-x-1=0 has:
View answer and explanation
Correct option: A
Only one real root x=0x=0
Option A (Only one real root x=0) is correct because the exponential function exe^x is equal to x+1x+1 only at x=0x=0.
837 Easy
837 of 914
Application +4 | -1
The solution set of the inequation ∣3x+1∣>2,\left| {\frac{3}{x} + 1} \right| > 2, is
View answer and explanation
Correct option: C
(−1,  0)∪(0,  3)\left( { - 1,\;0} \right) \cup \left( {0,\;3} \right)
Option C ((−1,  0)∪(0,  3)\left( { - 1,\;0} \right) \cup \left( {0,\;3} \right)) is correct because it represents the solution set where the absolute value of the given expression is greater than 2.
838 Medium
838 of 914
Analyzing +4 | -1
The value of kk for which the equation (k−2)x2+8x+k+4=0\left( {k - 2} \right){x^2} + 8x + k + 4 = 0 has both roots real, distinct and negative, is
View answer and explanation
Correct option: C
3
Option C (3) is correct because it satisfies the conditions for the quadratic equation to have real, distinct, and negative roots.
839 Easy
839 of 914
Application +4 | -1
If a+b≠0a+b \neq 0 and the roots of x2−px+q=0x^2-p x+q=0 differ by -1 , then p2−4qp^2-4 q equals:
View answer and explanation
Correct option: C
1
Option C (1) is correct because the difference of roots being -1 allows us to relate the sum and product of roots, leading to the value of p2 - 4q.
840 Hard
840 of 914
Analyzing +4 | -1
Let a,b,ca,b,c be real number a≠0. If  αa \ne 0.\,\mathrm{If}\;\alpha is a root of a2x2+bx+c=0,  β{a^2}{x^2} + bx + c = 0,\;\beta is a root of a2x2−bx−c=0{a^2}{x^2} - bx - c = 0 and 0<α<β0 < \alpha < \beta, then the equation a2x2+2bx+2c=0{a^2}{x^2} + 2bx + 2c = 0 has a root of γ\gamma that always satisfies
View answer and explanation
Correct option: D
α<γ<β\alpha < \gamma < \beta
Option D (α<γ<β\alpha < \gamma < \beta) is correct because by Intermediate Value Theorem, since f(α) and f(β) have opposite signs, there must be a root between α and β.
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