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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Easy

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Showing 121-150 of 528 questions

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121 Easy
121 of 528
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.
122 Easy
122 of 528
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.
123 Easy
123 of 528
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.
124 Easy
124 of 528
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.
125 Easy
125 of 528
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.
126 Easy
126 of 528
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.
127 Easy
127 of 528
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.
128 Easy
128 of 528
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.
129 Easy
129 of 528
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.
130 Easy
130 of 528
Analyzing +4 | -1
If the equation 2x2−7x+1=02{x^2} - 7x + 1 = 0 and ax2+bx+2=0a{x^2} + bx + 2 = 0 have a common root, then
View answer and explanation
Correct option: C
a=4,b=−14a = 4,b = - 14
Option C (a=4,b=−14a = 4, b = - 14) is correct because it satisfies the condition for the two quadratic equations to have a common root.
131 Easy
131 of 528
Application +4 | -1
The number of real solutions of the equation,&nbsp;x2−∣x∣−12=0{x^2} - \left| x \right| - 12 = 0 is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the equation has two real solutions, obtained by solving for |x| and then for x.
132 Easy
132 of 528
Application +4 | -1
The number which exceeds its positive square root by 12 is:
View answer and explanation
Correct option: B
16
Option B (16) is correct because 16 exceeds its square root (4) by 12.
133 Easy
133 of 528
Application +4 | -1
Let α,  α2\alpha ,\;{\alpha ^2} be the roots of x2+x+1=0{x^2} + x + 1 = 0, then the equation whose roots are α31,  α62{\alpha ^{31}},\;{\alpha ^{62}} is
View answer and explanation
Correct option: C
x2+x+1=0{x^2} + x + 1 = 0
Option C (x2+x+1=0{x^2} + x + 1 = 0) is correct because it is derived by using the properties of cube roots of unity and substituting the values of the roots.
134 Easy
134 of 528
Application +4 | -1
If the difference between the roots of the equation x2+ax+1=0{x^2} + ax + 1 = 0 is less than 5\sqrt 5, then the set of possible values of aa is
View answer and explanation
Correct option: A
(−3,3)\left( { - 3,\mathrm{ }3} \right)
Option A ((−3,3)\left( { - 3,\mathrm{ }3} \right)) is correct because by finding the difference between the roots and using the given condition, we can determine the range of possible values for aa.
135 Easy
135 of 528
Application +4 | -1
cos⁡ec 18∘\cos ec\,18^\circ is a root of the equation
View answer and explanation
Correct option: D
x2−2x−4 = 0{x^\mathrm{2}} - 2x - 4\, = \,0
Option D (x2−2x−4 = 0{x^\mathrm{2}} - 2x - 4\, = \,0) is correct because cos⁡ec 18∘\cos ec\,18^\circ simplifies to 5+1\sqrt 5 + 1, which satisfies the given quadratic equation.

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136 Easy
136 of 528
Application +4 | -1
If α,β,γ\alpha, \beta, \gamma are the roots of 2x3+x2−7=02 x^{3}+x^{2}-7=0, then find the value of ∑(αβ+βα)\sum\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)
View answer and explanation
Correct option: B
-3
Option B (-3) is correct because by Vieta's formulas and algebraic manipulation, the expression simplifies to -3.
137 Easy
137 of 528
Application +4 | -1
Suppose aa and bb are rational and α\alpha and β\beta are the roots of x2+2ax+b=0x^2+2 a x+b=0, then the equation with rational coefficients with one root α+βα2+β2\alpha+\beta \sqrt{\alpha^2+\beta^2} is
View answer and explanation
Correct option: A
x2+4ax+2b=0x^2+4 a x+2 b=0
Option A (x2+4ax+2b=0x^2+4 a x+2 b=0) is correct because it is derived by finding the sum and product of the roots, where one root is given as α+βα2+β2\alpha+\beta \sqrt{\alpha^2+\beta^2}.
138 Easy
138 of 528
Application +4 | -1
If one root of a quadratic equation is −i+32\frac{-i+\sqrt{3}}{2}, then the sum of the roots is
View answer and explanation
Correct option: D
3\sqrt{3}
Option D (3\sqrt{3}) is correct because if one root is 3−i2\frac{{\sqrt{3}-i}}{{2}}, the other root is its conjugate 3+i2\frac{{\sqrt{3}+i}}{{2}}, and their sum is 3\sqrt{3}.
139 Easy
139 of 528
Application +4 | -1
If the roots of 4x2+5k=(5k+1)x4 x^2+5 k=(5 k+1) x differ by unity, then the negative value of kk is
View answer and explanation
Correct option: C
−15-\frac{1}{5}
Option C (-15\frac{1}{5}) is correct because by setting up equations for the sum and product of roots that differ by unity, we can solve for kk and find its negative value.
140 Easy
140 of 528
Application +4 | -1
If pp and qq are the roots of the equation x2+px+q=0x^{2}+p x+q=0, then
View answer and explanation
Correct option: A
p=1,q=−2p=1, q=-2
Option A (p=1,q=−2p=1, q=-2) is correct because substituting these values into the equation x2+px+q=0x^{2}+p x+q=0 shows that p and q are indeed the roots.
141 Easy
141 of 528
Application +4 | -1
Let f(x)=x2−2x+4x2+2x+4,x∈Rf(x)=\frac{x^2-2 x+4}{x^2+2 x+4}, x \in \mathbf{R}. The range of f(x)f(x) is
View answer and explanation
Correct option: B
[1/3,3][1 / 3,3]
Option B ([1/3,3][1 / 3,3]) is correct because finding the range of the function involves setting up an inequality and solving for the possible values of y.
142 Easy
142 of 528
Application +4 | -1
Product of roots of the equation x−3x−6=2x-\sqrt{3 x-6}=2 is
View answer and explanation
Correct option: D
10
Option D (10) is correct because the product of the roots of the equation is 10.
143 Easy
143 of 528
Analyzing +4 | -1
The quadratic equation x2−6x+a=0x^2-6 x+a=0 and x2−cx+6=0x^2-c x+6=0 have one root in common. The other roots of the first and second equations are integers in the ratio 4:34: 3. The common root is
View answer and explanation
Correct option: C
2
Option C (2) is correct because it is the only common root that satisfies the given conditions, where the other roots are integers in the ratio 4:3.
144 Easy
144 of 528
Analyzing +4 | -1
If every pair of the equations x2+px+qr=0x^2+p x+q r=0, x2+qx+rp=0,x2+rx+pq=0x^2+q x+r p=0, \quad x^2+r x+p q=0 have a common root, then the sum of three common roots is:
View answer and explanation
Correct option: A
−(p+q+r)2\frac{-(p+q+r)}{2}
Option A (−(p+q+r)2\frac{-(p+q+r)}{2}) is correct because by summing the roots of each equation and solving the resulting system of equations, we find that the sum of the three common roots is −(p+q+r)2\frac{-(p+q+r)}{2}.
145 Easy
145 of 528
Application +4 | -1
If two roots of the equation x3−px2+qx−r=0,(r≠0)x^3-p x^2+q x-r=0,(r \neq 0) are equal in magnitude but opposite in sign, then:
View answer and explanation
Correct option: C
pq=rp q=r
Option C (pq=rp q=r) is correct because if two roots are equal in magnitude but opposite in sign, then substituting the roots into Vieta's formulas leads to the relationship pq=rpq = r.
146 Easy
146 of 528
Analyzing +4 | -1
The value of ' aa ' for which the equation x3+ax+1=0x^3+a x+1=0 and x4+ax2+1=0x^4+a x^2+1=0, have a common root, is
View answer and explanation
Correct option: B
a=−2a=-2
Option B (a=−2a=-2) is correct because by finding the common root of the two equations and substituting it back, we get a = -2.
147 Easy
147 of 528
Application +4 | -1
If α,β\alpha, \beta are roots x2+px+q=0x^2+p x+q=0, then value of α3+β3\alpha^3+\beta^3 is
View answer and explanation
Correct option: B
3pq−p33 p q-p^3
Option B (3pq−p33pq-p^3) is correct because it is the correct application of the identity for α3+β3\alpha^3 + \beta^3 using Vieta's formulas.
148 Easy
148 of 528
Application +4 | -1
The set of values of aa for which 1 lies between the roots of equation x2−ax−a+3=0x^{2}-a x-a+3=0 is
View answer and explanation
Correct option: D
(2,∞)(2, \infty)
The correct answer is D ((2,∞)(2, \infty)) because for 1 to lie between the roots, f(1) must be less than 0, leading to the inequality a > 2.
149 Easy
149 of 528
Application +4 | -1
The values of xx and yy such that yy satisfy the equation (x,  y∈(x,\;y \in real numbers) x2−xy+y2−4x−4y+16=0{x^2} - xy + {y^2} - 4x - 4y + 16 = 0
View answer and explanation
Correct option: A
4,44,4
Option A (4,4) is correct because the condition for real x leads to y=4, and substituting back gives x=4.
150 Easy
150 of 528
Application +4 | -1
If tan⁡25∘\tan 25^{\circ} and tan⁡20∘\tan 20^{\circ} are roots of the quadratic equation x2+2px+q=0x^2+2 p x+q=0, then 2p−q2 p-q is equal to
View answer and explanation
Correct option: B
-1
Option B (-1) is correct because using the tangent addition formula and Vieta's formulas, we can relate the sum and product of the roots to p and q, and then solve for 2p-q.
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