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

The roots of the equation $x^2-2\sqrt 2x+1=0$ are-

  1. Real and distinct

  2. Imaginary and different

  3. Real and equal

  4. Rational and different.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The discriminant D = b^2 - 4ac = (-2*sqrt(2))^2 - 4(1)(1) = 8 - 4 = 4. Since D > 0 and is a perfect square, the roots are real, distinct, and rational.

Multiple choice

The roots of the equation $(b+c)x^2-(a+b+c)x+a=0 ::: (a,b,c :\epsilon:Q,b+c \neq a)$ are:

  1. Irrational and different

  2. Rational and different

  3. Imaginary and different

  4. Real and equal
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The discriminant D = (a+b+c)^2 - 4(b+c)a = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 4ab - 4ac = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc = (a-b-c)^2 + 4bc. Since a,b,c are rational, the discriminant is a square of a rational number, so the roots are rational.

Multiple choice

Sum of the roots of the equation $(x+3)^2-4|x+3|+3=0$ is-
  1. $4$
  2. $12$
  3. $-12$
  4. $-4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let y = |x+3|. The equation is y^2 - 4y + 3 = 0. (y-3)(y-1) = 0. So |x+3| = 3 or |x+3| = 1. Case 1: x+3 = 3 or -3, so x = 0 or -6. Case 2: x+3 = 1 or -1, so x = -2 or -4. Sum = 0 - 6 - 2 - 4 = -12.

Multiple choice

The roots of the quadratic equation $(a+b-2c)x^2-(2a-b-c)x+(a-2b+c)=0$ are-
  1. $(a+b+c)$ and $(a-b-c)$
  2. $\dfrac 12$ and $a-2b+c$
  3. $a-2b+c$ and $\dfrac 1{(a+b-2c)}$

  4. None of the above.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let f(x) = (a+b-2c)x^2 - (2a-b-c)x + (a-2b+c). Evaluating f(1) gives (a+b-2c) - (2a-b-c) + (a-2b+c) = a-a+b+b-2b-2c+c+c = 0, so x=1 is a root. The product of roots is (a-2b+c)/(a+b-2c), so the other root is (a-2b+c)/(a+b-2c). Neither option A, B, nor C matches these values.

Multiple choice

If both the roots of the equation $x^2-9x+a=0$ are positive and one is greater than $3$ and other is less than $3$, then all possible values of $a$ is
  1. $0$ $<$ $a$ $<$ $18$
  2. $-1$ $<$ $a$ $<$ $2$
  3. $-18$ $<$ $a$ $<$ $0$

  4. None of the above.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the roots of x^2 - 9x + a = 0 to be on opposite sides of 3, the function f(x) = x^2 - 9x + a must satisfy f(3) < 0. Calculating f(3) = 9 - 27 + a = a - 18, so a - 18 < 0 implies a < 18. Since the roots are positive, the product of roots (a) must be > 0, so 0 < a < 18.

Multiple choice

If $\alpha, \beta$ are the roots of the quadratic equation $(p^2+p+1)x^2+(p-1)x+p^2=0$ such that unity lies between the roots then the set of values of $p$ is:
  1. $\phi$
  2. $p\:\:\epsilon\:\:(-\infty,-1)U(0,\infty)$
  3. $p\:\:\epsilon\:\: (-1,0)$
  4. $(-1,1)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For unity to lie between the roots of f(x) = ax^2 + bx + c, a * f(1) < 0. f(1) = (p^2+p+1) + (p-1) + p^2 = 2p^2 + 2p. So (p^2+p+1)(2p^2+2p) < 0. Since p^2+p+1 is always positive, 2p(p+1) < 0, which holds for p in (-1, 0).

Multiple choice

$x^2-(m-3)x+m=0::(m \in R)$ be a quadratic equation. Find the value of $m$ for which b oth the roots are greater than $2$
  1. $(9,10)$
  2. $(-5,4)$
  3. $(6,\infty)$

  4. None of the above.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For roots > 2: 1) D >= 0: (m-3)^2 - 4m >= 0 => m^2 - 6m + 9 - 4m >= 0 => m^2 - 10m + 9 >= 0 => (m-9)(m-1) >= 0. m <= 1 or m >= 9. 2) f(2) > 0: 4 - 2(m-3) + m > 0 => 4 - 2m + 6 + m > 0 => 10 - m > 0 => m < 10. 3) Vertex > 2: (m-3)/2 > 2 => m-3 > 4 => m > 7. Combining: m >= 9 and m < 10. So (9, 10).

Multiple choice

If $b$ and $c$ are odd integers, then the equation $x^2+bx+c=0$ has:

  1. two integers roots, one even and one odd

  2. no integers roots

  3. two odd roots

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If b and c are odd, the discriminant D = b^2 - 4c. Since b is odd, b^2 is odd. Since c is odd, 4c is even. D = odd - even = odd. For integer roots, D must be a perfect square. However, an odd perfect square is always congruent to 1 mod 4. Here, b^2 is 1 mod 4 (if b is odd) and 4c is 0 mod 4, so D is 1 mod 4. This doesn't rule out roots, but testing parity shows that if x is even, x^2+bx+c is odd+even+odd = even (not 0), and if x is odd, x^2+bx+c is odd+odd+odd = odd (not 0). Thus, no integer roots exist.

Multiple choice

The equation $x^2 + nx + m = 0,$ where $n, m \in I$ can not have:

  1. integral roots

  2. non-integral rational roots

  3. irrational roots

  4. complex roots
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For x^2 + nx + m = 0 with integer coefficients, if the roots are rational, they must be integers (by the Rational Root Theorem). Thus, it cannot have non-integral rational roots.

Multiple choice

The least integral value of $a$ for which the equation $x^2-2(a-1)x+2a+1=0$ has both the roots positive is-
  1. $3$
  2. $4$
  3. $1$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For roots to be positive: 1) Discriminant D >= 0: 4(a-1)^2 - 4(2a+1) >= 0 => a^2 - 2a + 1 - 2a - 1 >= 0 => a^2 - 4a >= 0 => a(a-4) >= 0. So a <= 0 or a >= 4. 2) Sum of roots > 0: 2(a-1) > 0 => a > 1. 3) Product of roots > 0: 2a+1 > 0 => a > -0.5. Combining these, a >= 4. The least integral value is 4.