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

If $\alpha, \beta$ are the roots of equation $x^2 + 2(a - 3)x + 9 = 0$, $a\in R$ and $\alpha < 6 < \beta$, then $a <-\displaystyle{\frac{3}{4}}$.

  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
A Correct answer
Multiple choice

The values of $a$ for which the equation $3x^2 + 2(a^2 -3a + 2) = 0$ will have roots of opposite sign lie in the interval
  1. $(-\infty, 1)$
  2. $(-\infty, 0)$
  3. $(1, 2)$
  4. $\displaystyle{\left( \frac{3}{2}, 2 \right)}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the quadratic equation 3x^2 + 2(a^2 - 3a + 2) = 0 to have roots of opposite signs, the product of the roots (c/a) must be negative. Here, c = 2(a^2 - 3a + 2) and a = 3. So, 2(a^2 - 3a + 2) / 3 < 0, which means a^2 - 3a + 2 < 0. Factoring gives (a-1)(a-2) < 0. This inequality holds when 1 < a < 2.

Multiple choice

If $a + b + c = 0$ and $a, b, c$ are rational, then the roots of the equation $(b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0$ are rational .

  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
A Correct answer
Multiple choice

If $\displaystyle a,\alpha _{1},\alpha _{2},...,\alpha _{2n-1},b$ are in AP $\displaystyle a,\beta _{1},\beta _{2},...,\beta _{2n-1},b $ in GP and a, $\displaystyle a,\gamma _{1},\gamma _{2},...,\gamma _{2n-1},b$ are in HP where a, b are positive then the equation $\displaystyle \alpha _{n}x^{2}-\beta _{n}x+\gamma _{n}=0$ has

  1. real and equal roots

  2. real and unequal roots

  3. imaginary roots

  4. roots which are in AP
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In AP, GP, HP sequences with same endpoints, the middle terms satisfy alpha_n * gamma_n = beta_n^2. The discriminant of alpha_n x^2 - beta_n x + gamma_n = 0 is D = beta_n^2 - 4 * alpha_n * gamma_n. Since beta_n^2 = alpha_n * gamma_n, D = -3 * alpha_n * gamma_n. Since a, b are positive, all terms are positive, so D < 0, implying imaginary roots.

Multiple choice

Find the value of $k$ for which the quadratic equation $\displaystyle \left ( k-2 \right )x^{2}+2\left ( 2k-3 \right )x+5k-6=0$ has equal roots

  1. 1

  2. 3

  3. A and B both

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

For equal roots, the discriminant D = b^2 - 4ac = 0. Here a = k-2, b = 2(2k-3), c = 5k-6. D = 4(2k-3)^2 - 4(k-2)(5k-6) = 0. Simplifying: (4k^2 - 12k + 9) - (5k^2 - 16k + 12) = 0, which leads to -k^2 + 4k - 3 = 0, or k^2 - 4k + 3 = 0. Roots are k=1 and k=3.

Multiple choice

If the sum of the roots of the quadratic equation $\displaystyle ax^{2}+bx+c=0 $ is equal to sum of the squares of their reciprocals then $\displaystyle \frac{a}{c},\frac{b}{a},\frac{c}{b} $ are in

  1. AP

  2. GP

  3. HP

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

Sum of roots = -b/a. Sum of squares of reciprocals = (b^2 - 2ac)/c^2. Equating: -b/a = (b^2 - 2ac)/c^2. This leads to -bc^2 = ab^2 - 2a^2c. Dividing by abc: -c/a = b/c - 2a/b. Rearranging: 2a/b = b/c + c/a. This is the condition for HP.

Multiple choice

If the roots of the equation $\displaystyle \left ( a^{2}+b^{2} \right )x^{2}-2b\left ( a+c \right )x+\left ( b^{2}+c^{2} \right )=0 $ are equal then
  1. $2b = ac$
  2. $\displaystyle b^{2}=ac $
  3. $\displaystyle b=\frac{2ac}{a+c} $

  4. b = ac
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For equal roots, the discriminant D = b^2 - 4ac must be zero. Applying this to the given equation: (-2b(a+c))^2 - 4(a^2+b^2)(b^2+c^2) = 0. Expanding and simplifying leads to 4b^2(a^2+2ac+c^2) - 4(a^2b^2 + a^2c^2 + b^4 + b^2c^2) = 0, which reduces to b^2 = ac.

Multiple choice

If $\displaystyle a, b \, \varepsilon \, R, \ a\neq 0$ and the quadratic equation $\displaystyle ax^{2}-bx+1=0$ has imaginary roots, then $a + b +1$ is

  1. positive

  2. negative

  3. zero
  4. depends on the sign of $b$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For ax^2 - bx + 1 = 0 to have imaginary roots, the discriminant D = b^2 - 4a < 0, so b^2 < 4a. Since a != 0 and b^2 >= 0, a must be positive. For x=1, the value is a - b + 1. This is not directly helpful. Let's check the vertex or specific values. If roots are imaginary, the parabola ax^2 - bx + 1 does not cross the x-axis. Since a > 0, the parabola opens upward and is always above the x-axis. Thus, for any x, ax^2 - bx + 1 > 0. At x = -1, a + b + 1 > 0.