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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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Showing 421-450 of 914 questions

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421 Easy
421 of 914
Analyzing +4 | -1
If the roots of the equation qx2+px+q=0q{x^2} + px + q = 0 are complex, where p,qp,q are real, then the roots of the equation x2−4qx+p2=0{x^2} - 4qx + {p^2} = 0 are
View answer and explanation
Correct option: A
Real and unequal
Option A (Real and unequal) is correct because the discriminant of the second equation is positive, indicating real and unequal roots.
422 Easy
422 of 914
Application +4 | -1
If the product of the roots of the equation x2−5kx+2e4hx−1=0x^2-5 k x+2 e^{4 h x}-1=0 is 31 , then sum of the root is
View answer and explanation
Correct option: D
10
Option D (10) is correct because solving for k from the product of roots and then finding the sum of the roots gives 10.
423 Medium
423 of 914
Application +4 | -1
If p,  q,  rp,\;q,\;r are in AP and are positive, the roots of the quadratic equation px2+qx+r=0p{x^2} + qx + r = 0 are all real for
View answer and explanation
Correct option: A
∣rp−7∣≥43\left| {\frac{r}{p} - 7} \right| \ge 4\sqrt 3
Option A (∣rp−7∣≥43\left| {\frac{r}{p} - 7} \right| \ge 4\sqrt 3) is correct because it represents the condition for the roots of the quadratic equation to be real, derived from the given AP condition.
424 Easy
424 of 914
Analyzing +4 | -1
If x2+3x+5=0x^2+3 x+5=0 and ax2+bx+c=0a x^2+b x+c=0 have aa common root and a,b,c∈Na, b, c \in \mathcal{N}, the minimum value of a+b+ca+b+c is
View answer and explanation
Correct option: B
9
Option B (9) is correct because the minimum value of a+b+c is obtained when the two quadratic equations have both roots in common and a, b, c are natural numbers.
425 Easy
425 of 914
Application +4 | -1
Suppose a∈Ia \in \mathbf{I} and the equation (x−a)(x−5)=3(x-a)(x-5)=3 has integral roots, then the set of values which aa can take is:
View answer and explanation
Correct option: C
{3,7}\{3,7\}
Option C ({3,7}\{3,7\}) is correct because these are the only integer values of 'a' that result in integer roots for the given equation.
426 Medium
426 of 914
Analyzing +4 | -1
If α\alpha and β\beta are the roots of ax2+bx+c=0a{x^2} + bx + c = 0, then the equation ax2−bx(x−1)+c(x−1)2=0a{x^2} - bx\left( {x - 1} \right) + c{\left( {x - 1} \right)^2} = 0 has roots
View answer and explanation
Correct option: C
αα+1,ββ+1\frac{\alpha }{{\alpha + 1}},\frac{\beta }{{\beta + 1}}
Option C (αα+1,ββ+1\frac{\alpha }{{\alpha + 1}},\frac{\beta }{{\beta + 1}}) is correct because substituting x=y1+yx = \frac{y}{{1+y}} into the transformed equation gives the original equation with roots α and β.
427 Easy
427 of 914
Application +4 | -1
If α,β\alpha ,\beta and γ\gamma are the roots of x2+qx+r=0,{x^2} + qx + r = 0, then ∑αβ+γ,\sum \frac{\alpha }{{\beta + \gamma }}, is
View answer and explanation
Correct option: D
−3- 3
Option D (-3) is correct because given the roots and the equation, the expression simplifies to -3 using the relationship between roots and coefficients.
428 Easy
428 of 914
Analyzing +4 | -1
f(x)f(x) is a cubic polynomial x3+ax2+bx+cx^3+a x^2+b x+c such that f(x)=0f(x)=0 has three distinct integral roots and f(g(x))=0f(g(x))=0 does not have real roots, where g(x)=x2+2x−5g(x)=x^2+2 x-5, the minimum value of a+b+ca+b+c is
View answer and explanation
Correct option: C
719
The correct answer is 719 because it is the minimum value of a+b+c that satisfies the given conditions for the cubic polynomial f(x).
429 Medium
429 of 914
Application +4 | -1
Let a,ba, b and cc be real numbers with a≠0a \neq 0 and let α,β\alpha, \beta be the roots of the equation ax2+bx+c=0a x^2+b x+c=0. Then, a3x2+abcx+c3=0a^3 x^2+a b c x+c^3=0 has roots
View answer and explanation
Correct option: A
α2β,β2α\alpha^2 \beta, \beta^2 \alpha
Option A (α2β,β2α\alpha^2 \beta, \beta^2 \alpha) is correct because these roots are derived by substituting xx with ca\frac{{c}}{{a}} times the original roots in the transformed equation.
430 Medium
430 of 914
Analyzing +4 | -1
If a,ba,b are distinct positive real numbers, then which one of the following is true?
View answer and explanation
Correct option: A
a4+b4>a3b+ab3{a^4} + {b^4} > {a^3}b + a{b^3}
Option A (a4+b4>a3b+ab3{a^4} + {b^4} > {a^3}b + a{b^3}) is correct because the expression a4+b4−a3b−ab3{a^4} + {b^4} - {a^3}b - a{b^3} can be factored into a positive quantity.
431 Easy
431 of 914
Application +4 | -1
The number of values of a for which (a2−3a+2)x2+(a2−5a+6)x+a2−4=0\left( {{a^2} - 3a + 2} \right){x^2} + \left( {{a^2} - 5a + 6} \right)x + {a^2} - 4 = 0 is an identity in xx, is
View answer and explanation
Correct option: C
1
Option C (1) is correct because a=2 is the only value that simultaneously satisfies all three equations derived from the coefficients of the given identity.
432 Medium
432 of 914
Analyzing +4 | -1
If the sum of the roots of the quadratic equation ax2+bx+c=0a x^2+b x+c=0 is equal to the sum of the squares of their reciprocals, then a/c,b/a,c/ba / c, b / a, c / b are in:
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the given condition leads to a relationship where the reciprocals are in A.P., thus the original terms are in H.P.
433 Medium
433 of 914
Analyzing +4 | -1
If cc and dd are roots of the equation (x−a)(x−b)−k=0,\left( {x - a} \right)\left( {x - b} \right) - k = 0, then a,ba,b are roots of the equation
View answer and explanation
Correct option: B
(x−c)(x−d)+k=0\left( {x - c} \right)\left( {x - d} \right) + k = 0
Option A (x−c)(x−d)−k=0\left( {x - c} \right)\left( {x - d} \right) - k = 0 is incorrect. The correct answer is B. Option B (x−c)(x−d)+k=0\left( {x - c} \right)\left( {x - d} \right) + k = 0 is correct because it is derived by substituting x=a and x=b into the equation and rearranging.
434 Easy
434 of 914
Analyzing +4 | -1
If f(x)f(x) is a quadratic expression such that f(a)+f(b)=0f(\mathrm{a})+f(\mathrm{b})=0, and -1 is a root of f(x)=0f(x)=0, then the other root of f(x)=0f(x)=0 is
View answer and explanation
Correct option: D
85\frac{8}{5}
Option D (85\frac{8}{5}) is correct because by using the given condition f(a)+f(b)=0f(a)+f(b)=0 and the fact that -1 is a root, we can find the other root of the quadratic expression.
435 Medium
435 of 914
Application +4 | -1
The sum of all the real roots of the equation ∣x−2∣2+∣x−2∣−2=0{\left| {x - 2} \right|^2} + \left| {x - 2} \right| - 2 = 0 is
View answer and explanation
Correct option: D
4
The correct answer is D (4) because the sum of the real roots obtained by solving the given equation for both cases (x >= 2 and x < 2) is 4.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
436 Easy
436 of 914
Application +4 | -1
If 2+i32 + i\sqrt 3 is a root of x2+px+q=0{x^2} + px + q = 0 where p,q∈R,p,q \in R,, then
View answer and explanation
Correct option: A
p=−4,q=7p = - 4,q = 7
Option A (p=−4,q=7p = - 4,q = 7) is correct because the sum and product of the roots of the quadratic equation are used to find the values of p and q, considering the complex conjugate root.
437 Easy
437 of 914
Application +4 | -1
If 2−i2-i is a root of the equation ax2+12x+b=0a x^2+12 x+b=0 (where aa and bb are real), then the value of aba b is equal to
View answer and explanation
Correct option: A
45
Option A (45) is correct because by finding the other root and using the relationship between roots and coefficients, we can determine the values of a and b and then calculate ab.
438 Medium
438 of 914
Analyzing +4 | -1
If both the roots of the quadratic equation x2−2kx+k2+k−5=0{x^2} - 2kx + {k^2} + k - 5 = 0 are less than 5, then kk lies in the interval
View answer and explanation
Correct option: B
(−∞,  4)\left( { - \infty ,\;4} \right)
Option B ((−∞,;4)\left( { - \infty ,\\;4} \right)) is correct because the roots of the quadratic equation are less than 5 only when k is less than 4.
439 Easy
439 of 914
Analyzing +4 | -1
The solution of set of the equation xlog⁡x(1−x)2=9{x^{{{\log }_x}{{\left( {1 - x} \right)}^2}}} = 9 is
View answer and explanation
Correct option: B
{  4}\left\{ {\;4} \right\}
Option B ({4}\{4\}) is correct because only 4 satisfies the given equation and the domain of the logarithmic function.
440 Medium
440 of 914
Analyzing +4 | -1
Solution set of log⁡3(x2−2)<log⁡3(32∣x∣−1)\log _3\left(x^2-2\right)<\log _3\left(\frac{3}{2}|x|-1\right) is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the solution set obtained by solving the logarithmic inequality does not match any of the given options.
441 Easy
441 of 914
Application +4 | -1
If x2/3−7x1/3+10=0{x^{2/3}} - 7{x^{1/3}} + 10 = 0, then the value of xx is
View answer and explanation
Correct option: D
{125,8}\left\{ {125,8} \right\}
Option D ({125,8}\left\{ {125,8} \right\}) is correct because both 125 and 8 are solutions to the given equation after substituting y=x1/3y = x^{1/3} and solving the resulting quadratic equation.
442 Medium
442 of 914
Application +4 | -1
In a triangle ABC,∠C=π2\mathrm{ABC}, \angle \mathrm{C}=\frac{\pi}{2}. If tan⁡(A2)\tan \left(\frac{\mathrm{A}}{2}\right) and tan⁡(B2)\tan \left(\frac{\mathrm{B}}{2}\right) are the roots of the equation ax2+bx+c=0(a≠0)a x^2+b x+c=0(a \neq 0) then&nbsp;
View answer and explanation
Correct option: A
a+b=c\mathrm{a}+\mathrm{b}=\mathrm{c}
Option A (a+b=c) is correct because by using the given condition ∠ C = π2\frac{π}{2} and the properties of tangent function, we can derive the relationship a+b=c.
443 Medium
443 of 914
Application +4 | -1
The solution set of ∣3−4x∣≥9|3-4 x| \geq 9 is
View answer and explanation
Correct option: B
(−∞,−32]∪[3,∞)\left(-\infty,-\frac{3}{2}\right] \cup[3, \infty)
Option B ((−∞,−32]∪[3,∞)\left(-\infty,-\frac{{3}}{{2}}\right] \cup[3, \infty)) is correct because it represents the solution set where the absolute value expression is greater than or equal to 9.
444 Medium
444 of 914
Analyzing +4 | -1
The system y(x2+7x+12)=1   and  x+y=6,y>0{y^{\left( {{x^2} + 7x + 12} \right)}} = 1\;\,\mathrm{and}\;x + y = 6,y > 0 has
View answer and explanation
Correct option: D
More than 2 solutions
Option D (More than 2 solutions) is correct because by solving the equations y(x2+7x+12)=1{y^{\left( {{x^2} + 7x + 12} \right)}} = 1 and x+y=6x + y = 6, we find three distinct solutions.
445 Medium
445 of 914
Application +4 | -1
If the sum of the roots of ax2+bx+c=0a x^2+b x+c=0 is equal to the sum of the squares, then
View answer and explanation
Correct option: A
2ac=ab+b22 a c=a b+b^2
Option A (2ac = ab + b²) is correct because it is derived from equating the sum of roots to the sum of squares of roots.
446 Medium
446 of 914
Analyzing +4 | -1
The number of real solutions of the equation 6−xx2−4=2+x(x+2)\frac{6-x}{x^2-4}=2+\frac{x}{(x+2)} is
View answer and explanation
Correct option: B
one
Option B (one) is correct because after simplifying the equation and excluding the extraneous solution, only one real solution remains.
447 Easy
447 of 914
Application +4 | -1
If 2−i2 - i is the root of the equation ax2+12x+b=0a{x^2} + 12x + b = 0 (where aa and bb are real), then the value of abab is equal to
View answer and explanation
Correct option: A
45
The correct answer is A (45) because by finding the conjugate root and using Vieta's formulas, we can determine the values of a and b, and their product is 45.
448 Easy
448 of 914
Application +4 | -1
If the roots of the quadratic equation ax2−5x+6=0a x^2-5 x+6=0 are in the ratio 2:32: 3, then ' aa ' is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because the value of 'a' is found by using the ratio of roots and solving the resulting equation.
449 Easy
449 of 914
Application +4 | -1
The product of the roots of the equation 9x2−18∣x∣+5=0,9{x^2} - 18|x| + 5 = 0, is
View answer and explanation
Correct option: D
2581\frac{{25}}{{81}}
The correct answer is D (2581\frac{{25}}{{81}}) because it is the product of the roots of the given equation, found by solving for |x| and then finding the roots.
450 Easy
450 of 914
Analyzing +4 | -1
The number of real solutions of the equation e∣x∣−∣x∣=0{e^{\left| x \right|}} - \left| x \right| = 0, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the exponential function e∣x∣{e^{\left| x \right|}} is always greater than ∣x∣\left| x \right|, so they never intersect.
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