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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Easy

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Showing 481-510 of 528 questions

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481 Easy
481 of 528
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.
482 Easy
482 of 528
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.
483 Easy
483 of 528
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.
484 Easy
484 of 528
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.
485 Easy
485 of 528
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.
486 Easy
486 of 528
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.
487 Easy
487 of 528
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.
488 Easy
488 of 528
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.
489 Easy
489 of 528
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.
490 Easy
490 of 528
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.
491 Easy
491 of 528
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.
492 Easy
492 of 528
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.
493 Easy
493 of 528
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.
494 Easy
494 of 528
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.
495 Easy
495 of 528
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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496 Easy
496 of 528
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.
497 Easy
497 of 528
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).
498 Easy
498 of 528
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.
499 Easy
499 of 528
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.
500 Easy
500 of 528
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.
501 Easy
501 of 528
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.
502 Easy
502 of 528
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.
503 Easy
503 of 528
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.
504 Easy
504 of 528
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.
505 Easy
505 of 528
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.
506 Easy
506 of 528
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.
507 Easy
507 of 528
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.
508 Easy
508 of 528
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.
509 Easy
509 of 528
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.
510 Easy
510 of 528
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.
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