Quantitative Aptitude

Algebra

3,715 Questions
All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice

Find the equation whose roots are $\displaystyle \left ( \alpha +\beta \right )^{2}:and:\left ( \alpha -\beta \right )^{2},$ where $\displaystyle \alpha$ and $\beta $ are the roots of $\displaystyle 2x^{2}+2\left ( m+n \right )x+\left ( m^{2}+n^{2} \right )= 0.$
  1. $\displaystyle x^{2}-4mnx-\left ( m^{2}+n^{2} \right )^{2}= 0.$
  2. $\displaystyle x^{2}+2mnx-\left ( m^{2}-n^{2} \right )^{2}= 0.$
  3. $\displaystyle x^{2}-4mnx-\left ( m^{2}-n^{2} \right )^{2}= 0.$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

A quadratic polynomial maps from $[-2,3]$ onto $[0,3]$ and touches x-axis at $x=3$,then the polynomial is
  1. $\displaystyle \frac{3}{16}(x^{2}-6x+16)$
  2. $\displaystyle \frac{3}{25}(x^{2}-6x+9)$
  3. $\displaystyle \frac{3}{25}(x^{2}-6x+16)$
  4. $\displaystyle \frac{3}{16}(x^{2}-6x+9)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the parabola touches the x-axis at x=3, it must be of the form a(x-3)^2. Evaluating at x=-2 gives a(-2-3)^2 = 25a. Since the range is [0,3], the maximum value at x=-2 is 3, so 25a = 3, a = 3/25. The polynomial is 3/25(x^2-6x+9).

Multiple choice

If the quadratic equation $ax^2 + bx + 6 = 0 $ does not have distinct real roots, then the least value of $2a + b$ is
  1. $2$
  2. $-3$
  3. $-6$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For no distinct real roots, the discriminant D = b^2 - 4ac <= 0. Here, b^2 - 4(a)(6) <= 0, so b^2 <= 24a. We want to minimize 2a + b. Let k = 2a + b, so b = k - 2a. Substituting into the inequality: (k - 2a)^2 <= 24a. For this to hold for some a, the quadratic in a must have a non-negative discriminant. Solving this leads to the minimum value -3.

Multiple choice

Find the values of $K$, so that the quadratic equation ${ x }^{ 2 }+2\left( K-1 \right) x+K+5=0$ has atleast one positive root.
  1. $K\le -2$
  2. $K\le 1$
  3. $K\le -1$
  4. $K\le 2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a quadratic equation to have at least one positive root, either the product of roots is negative (c/a < 0) or the discriminant is non-negative and the sum of roots is positive. Here, c/a = K+5. If K+5 < 0, K < -5. Testing the vertex and boundary conditions for the quadratic, the range K <= -1 ensures the existence of a positive root.

Multiple choice

In the following question, two equations numbered I and II are given. You have to solve both the equations and give answer if (I) $x^{2}+11x+28=0$ and (II) $ y^{2}+15y+56=0$
  1. $\displaystyle x>y$
  2. $\displaystyle x\geq y$
  3. $\displaystyle x< y$
  4. $\displaystyle x\leq y$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation I: x^2 + 11x + 28 = 0 => (x+4)(x+7) = 0, so x = -4, -7. Equation II: y^2 + 15y + 56 = 0 => (y+7)(y+8) = 0, so y = -7, -8. Comparing: -4 > -7, -4 > -8, -7 = -7, -7 > -8. Thus, x >= y.

Multiple choice

Statement-I : If roots of the equation $x^2- bx + c = 0$ are two consecutive integers, then $b^2- 4c = 1$. Statement-II : If $a, b, c$ are odd integers, then the roots of the equation $4abcx^2 + (b^2- 4ac)x- b = 0 $ are real and distinct.

  1. Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I

  2. Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I

  3. Statement-I is true, Statement-II is false

  4. Statement-I is false, Statement-II is true
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For Statement-I, if roots are n and n+1, the equation is (x-n)(x-n-1) = x^2 - (2n+1)x + n(n+1) = 0. Then b = 2n+1 and c = n^2+n. The discriminant b^2-4c = (2n+1)^2 - 4(n^2+n) = 4n^2+4n+1-4n^2-4n = 1. Statement-II is true because for odd a, b, c, the discriminant b^2-4ac is odd, ensuring real and distinct roots, but it does not explain Statement-I.

Multiple choice

Find the quadratic equation whose roots are half of the reciprocal of the roots of the equation $\displaystyle ax^{2}+bx+c=0$
  1. $\displaystyle 4ax^{2}+2bx+c=0$
  2. $\displaystyle 4cx^{2}+2bx+a=0$
  3. $\displaystyle 2cx^{2}+bx+a=0$
  4. $\displaystyle 2ax^{2}+bx+c=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If the roots of ax^2 + bx + c = 0 are r1 and r2, the new roots are 1/(2r1) and 1/(2r2). Let y = 1/(2x), so x = 1/(2y). Substituting this into the original equation gives a(1/(2y))^2 + b(1/(2y)) + c = 0, which simplifies to a/(4y^2) + b/(2y) + c = 0. Multiplying by 4y^2 yields a + 2by + 4cy^2 = 0.

Multiple choice

If p and q are the roots of the quadratic equation $\displaystyle x^{2}-\left ( \alpha -2 \right )x-\alpha -1=0$. What is the minimum possible value of $\displaystyle p^{2}+q^{2}$?

  1. 0

  2. 3

  3. 4

  4. 5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

p+q = alpha-2, pq = -alpha-1. p^2+q^2 = (p+q)^2 - 2pq = (alpha-2)^2 - 2(-alpha-1) = alpha^2 - 4alpha + 4 + 2alpha + 2 = alpha^2 - 2alpha + 6. Minimum value occurs at alpha = -(-2)/(2*1) = 1. Min value = 1^2 - 2(1) + 6 = 5.

Multiple choice

If $\displaystyle \sin a$ and $\cos a$ are the roots of the equation $\displaystyle 4x^{2}-kx-1=0,\left ( k> 0 \right )$ then the value of $k$ is
  1. $\displaystyle 2\sqrt{2}$
  2. $4$
  3. $2$
  4. $\displaystyle 4\sqrt{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Roots are sin(a) and cos(a). Sum = sin(a) + cos(a) = k/4. Product = sin(a)cos(a) = -1/4. (sin(a) + cos(a))^2 = 1 + 2sin(a)cos(a) = 1 + 2(-1/4) = 1 - 0.5 = 0.5. So (k/4)^2 = 0.5 => k^2/16 = 0.5 => k^2 = 8 => k = sqrt(8) = 2*sqrt(2).

Multiple choice

If the roots of the quadratic equation $\displaystyle 4x^{2}-kx-1=0$ are $\displaystyle \sin \theta $ and $\cos \theta $ , then $\displaystyle \left | k \right | $ has the value equal to
  1. $2$
  2. $8$
  3. $\displaystyle 4\sqrt{2}$
  4. $\displaystyle 2\sqrt{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For 4x^2 - kx - 1 = 0, the sum of roots sin(theta) + cos(theta) = k/4 and the product sin(theta)cos(theta) = -1/4. Squaring the sum: sin^2(theta) + cos^2(theta) + 2sin(theta)cos(theta) = k^2/16. Substituting 1 + 2(-1/4) = k^2/16 gives 1 - 1/2 = k^2/16, so 1/2 = k^2/16, k^2 = 8. Thus |k| = sqrt(8) = 2*sqrt(2).

Multiple choice

If A, G & H are respectively the AM, GM and HM of three positive numbers a, b & c then the equation whose roots are a, b & c is given by
  1. $\displaystyle x^{3}-3Ax^{2}+3G^{3}x-G^{3}=0 $
  2. $\displaystyle x^{3}-3Ax^{2}+3\left ( G^{3}/H \right )x-G^{3}=0 $
  3. $\displaystyle x^{3}+3Ax^{2}+3\left ( G^{3}/H \right )x-G^{3}=0 $
  4. $\displaystyle x^{3}-3Ax^{2}+3\left ( G^{3}/H \right )x+G^{3}=0 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For roots a, b, c, the equation is x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = 0. Given AM = (a+b+c)/3 = A, GM^3 = abc = G^3, and HM = 3abc/(ab+bc+ca) = H, we have ab+bc+ca = 3G^3/H. Substituting gives x^3 - 3Ax^2 + 3(G^3/H)x - G^3 = 0.