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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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Showing 241-270 of 914 questions

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241 Easy
241 of 914
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.
242 Easy
242 of 914
Application +4 | -1
The number of real solutions of the equation, 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.
243 Medium
243 of 914
Analyzing +4 | -1
Let f(x)=x2+ax+b;a,  b∈R.  If  f(1)+f(2)+f(3)=0,f\left( x \right) = {x^2} + ax + b;a,\;b \in R.\,\,\mathrm{If}\;f\left( 1 \right) + f\left( 2 \right) + f\left( 3 \right) = 0, then the roots of the equation f(x)=0f\left( x \right) = 0
View answer and explanation
Correct option: D
Real and distinct
Option D (Real and distinct) is correct. If f(1) + f(2) + f(3) = 0, the values of f(1), f(2), and f(3) cannot all have the same sign, implying the roots are real and distinct.
244 Hard
244 of 914
Analyzing +4 | -1
The number of negative integral solutions of x2.2x+1+2∣x−3∣+2=x2.2∣x−3∣+4+2x−1,{x^2}{.2^{x + 1}} + {2^{\left| {x - 3} \right| + 2}} = {x^2}{.2^{\left| {x - 3} \right| + 4}} + {2^{x - 1}}, is
View answer and explanation
Correct option: A
None
Option A (None) is correct because after analyzing the equation for both x≥3x \ge 3 and x<3x < 3, there are no negative integral solutions.
245 Medium
245 of 914
Application +4 | -1
If α,β\alpha, \beta be the roots of x2−px+q=0x^2-p x+q=0 and α′,β′\alpha^{\prime}, \beta^{\prime} be the roots of x2−p′x+q′=0x^2-p^{\prime} x+q^{\prime}=0, then the value of (α−α′)2+(β−α′)2+(a−β′)2+(β−β′)2\left(\alpha-\alpha^{\prime}\right)^2+\left(\beta-\alpha^{\prime}\right)^2+\left(a-\beta^{\prime}\right)^2+\left(\beta-\beta^{\prime}\right)^2 is:
View answer and explanation
Correct option: A
2{p2−2q+p′2−2q′−pp′}2\left\{p^2-2 q+p^{\prime 2}-2 q^{\prime}-p p^{\prime}\right\}
Option A (2{p2−2q+p′2−2q′−pp′}2\{p^2-2 q+p^{\prime 2}-2 q^{\prime}-p p^{\prime}\}) is correct because it is derived by expanding the given expression and substituting the sum and product of roots for both quadratic equations.
246 Easy
246 of 914
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.
247 Medium
247 of 914
Analyzing +4 | -1
The number of real solutions of the equation 3(x2+1x2)−2(x+1x)+5=03\left(x^{2}+\frac{1}{x^{2}}\right)-2\left(x+\frac{1}{x}\right)+5=0, is
View answer and explanation
Correct option: B
0
Option B (0) is correct because the resulting quadratic equation in 't' has no real solutions within the allowed range for t = x + 1/x.
248 Easy
248 of 914
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.
249 Medium
249 of 914
Application +4 | -1
If one root of the equation x2+px+12=0{x^2} + px + 12 = 0 is 4, while the equation x2+px+q=0{x^2} + px + q = 0 has equal roots, then the value of qq is
View answer and explanation
Correct option: D
494\frac{{49}}{4}
Option D (494\frac{49}{4}) is correct because by using the given root and the properties of quadratic equations, we can find the value of p and then use that to find q, which equals 494\frac{49}{4}.
250 Easy
250 of 914
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.
251 Easy
251 of 914
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.
252 Medium
252 of 914
Analyzing +4 | -1
Given, n4<10nn^4<10^n for a fixed positive integer n≥2n \geq 2, then
View answer and explanation
Correct option: A
(n+1)4<10n+1(n+1)^4 < 10^{n+1}
Option A ((n+1)4<10n+1(n+1)^4 < 10^{n+1}) is correct because it demonstrates that if n4<10nn^4 < 10^n for n≥2n \geq 2, then (n+1)4(n+1)^4 will also be less than 10n+110^{n+1}.
253 Medium
253 of 914
Analyzing +4 | -1
The solution set of x2+x+∣x∣+1<0,{x^2} + x + \left| x \right| + 1 < 0, is
View answer and explanation
Correct option: D
ϕ\phi
Option D (ϕ\phi) is correct because the inequality x2+x+∣x∣+1<0{x^2} + x + |x| + 1 < 0 has no real solutions for any value of x.
254 Easy
254 of 914
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.
255 Medium
255 of 914
Analyzing +4 | -1
x4−4x−1=0x^4-4 x-1=0
View answer and explanation
Correct option: C
exactly two real roots
Option C (exactly two real roots) is correct because the function f(x)=x4−4x−1f(x) = x^4 - 4x - 1 has only two real roots, one positive and one negative, as determined by its derivative and the function's values.

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256 Easy
256 of 914
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}.
257 Easy
257 of 914
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}.
258 Medium
258 of 914
Application +4 | -1
The value of 'cc' for which ∣α2−β2∣=7/4,\left| {{\alpha ^2} - {\beta ^2}} \right| = 7/4, where α\alpha and β\beta are the roots of 2  x2+7  x+c=0,2\;{x^2} + 7\;x + c = 0, is
View answer and explanation
Correct option: C
6
Option C (6) is the closest to the calculated value of 6, although the provided solution has an error. The correct value of c should be approximately 6.
259 Easy
259 of 914
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.
260 Medium
260 of 914
Analyzing +4 | -1
If x2+2x+2xy+my−3=0{x^2} + 2x + 2xy + my - 3 = 0 has two rational factors, then the values of mm will be
View answer and explanation
Correct option: C
6,  −26,\; - 2
Option C (6, -2) is correct because the given equation has two rational factors only when m = 6 or m = -2.
261 Easy
261 of 914
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.
262 Easy
262 of 914
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.
263 Medium
263 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of x2+bx+c=0x^2+b x+c=0 and are positive, then α+β\sqrt{\alpha}+\sqrt{\beta} is
View answer and explanation
Correct option: B
−b+2c\sqrt{-b+2 \sqrt{c}}
Option B (−b+2c\sqrt{{-b+2 \sqrt{{c}}}}) is correct because it is derived by using Vieta's formulas and algebraic manipulation.
264 Easy
264 of 914
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.
265 Easy
265 of 914
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.
266 Medium
266 of 914
Application +4 | -1
If 3p2=5p+23{p^2} = 5p + 2 and 3q2=5q+23{q^2} = 5q + 2 where p≠qp \ne q, then the equation whose roots are 3p−2q3p - 2q and 3q−2p3q - 2p is
View answer and explanation
Correct option: A
3x2−5x−100=03{x^2} - 5x - 100 = 0
Option A (3x2−5x−100=03{x^2} - 5x - 100 = 0) is correct because it is the quadratic equation formed with roots 3p−2q3p-2q and 3q−2p3q-2p derived from the given equations.
267 Easy
267 of 914
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}.
268 Easy
268 of 914
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.
269 Medium
269 of 914
Analyzing +4 | -1
If xx is real, the function (x−a)(x−b)(x−c)\frac{{\left( {x - a} \right)\left( {x - b} \right)}}{{\left( {x - c} \right)}} will assume all real values, provided
View answer and explanation
Correct option: D
a≤c≤ba \le c \le b
Option D (a≤c≤ba \le c \le b) is correct because it ensures the discriminant of the quadratic equation is non-negative for all real values of y, guaranteeing real roots for x.
270 Medium
270 of 914
Application +4 | -1
Let a,b,ca,b,c be positive real numbers and ax2+b/x2≥2a{x^2} + b/{x^2} \ge 2 for all x∈R+x \in {R^ + }. Then,
View answer and explanation
Correct option: A
4ab≥c24ab \ge {c^2}
Option A (4ab≥c24ab \ge {c^2}) is correct because it ensures that the given inequality holds for all positive real numbers x, derived from the minimum value of the expression.
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