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 $p$ and $q$ are odd integers, then the equation $x^{2}+2px+2q=0$

  1. has no integral root

  2. has no rational root

  3. has no irrational

  4. has no imaginary root.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Discriminant D = (2p)^2 - 4(1)(2q) = 4p^2 - 8q = 4(p^2 - 2q). For roots to be rational, D must be a perfect square. Since p is odd, p^2 is odd. 2q is even. p^2 - 2q is odd. 4 * (odd) is never a perfect square (it is 4 times an odd number). Thus, roots cannot be rational.

Multiple choice

The value of $p$ for which both the roots of the equation $4x^{2}-20px+\left(25p^{2}+15p-66\right)=0$ are less than $2$, lies in the interval
  1. $\left(\ -1,\dfrac{-4}{5}\right)$
  2. $\left(-\infty, -1\right)$
  3. $\left(2, \infty\right)$
  4. $\left(\ - \infty, \dfrac{-4}{5}\right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For roots to be less than 2, the vertex -b/2a < 2, f(2) > 0, and discriminant D >= 0. Vertex: 20p/8 = 2.5p < 2 => p < 0.8. f(2) = 4(4) - 20p(2) + 25p^2 + 15p - 66 = 16 - 40p + 25p^2 + 15p - 66 = 25p^2 - 25p - 50 > 0 => p^2 - p - 2 > 0 => (p-2)(p+1) > 0. This holds for p > 2 or p < -1. Combining with p < 0.8, we get p < -1.

Multiple choice

Let p and q be the roots of the equation ${x^2} - 2x + A = 0,$ and let r and s be the roots of the equation ${x^2} - 18x + B = 0.$ If p < q < r < s are in arithmetic progression. The value of (A+B) equals
  1. $80$
  2. $77$
  3. $75$
  4. $74$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Roots p, q are for x^2 - 2x + A = 0 (p+q=2, pq=A). Roots r, s are for x^2 - 18x + B = 0 (r+s=18, rs=B). AP: p, q, r, s. Let common difference be d. q=p+d, r=p+2d, s=p+3d. (p+q)=2p+d=2. (r+s)=2p+5d=18. Subtracting: 4d=16, d=4. 2p+4=2, p=-1. Roots are -1, 3, 7, 11. A = pq = -3. B = rs = 77. A+B = 74.

Multiple choice

If $\alpha,\beta$ are the roots of $x^{2}+6x+9=0$, then the equation whose roots are $\dfrac{1}{\alpha}, \dfrac{1}{\beta}$ is
  1. $2x^{2}+3x+18=0$
  2. $x^{2}+6x-9=0$
  3. $9x^{2}+6x+1=0$
  4. $9x^{2}-6x+1=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the equation x^2 + 6x + 9 = 0, the roots are alpha = -3 and beta = -3. The new roots are 1/alpha = -1/3 and 1/beta = -1/3, so the equation is (x + 1/3)(x + 1/3) = 0, which is x^2 + (2/3)x + 1/9 = 0, or 9x^2 + 6x + 1 = 0.

Multiple choice

Let $\alpha$ and $\beta$ be the roots of the equation $3x^{2}-6x+5=0$ then the equation whose roots are $(\alpha+\beta)$ and $\dfrac{2}{(\alpha+\beta)}$ is
  1. $x^{2}+3x-1=0$
  2. $x^{2}+3x-2=0$
  3. $x^{2}-3x+2=0$
  4. $x^{2}-3x-2=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For 3x^2 - 6x + 5 = 0, sum of roots (alpha + beta) = -(-6)/3 = 2. Product of roots (alpha * beta) = 5/3. The new roots are (alpha + beta) = 2 and 2/(alpha + beta) = 2/2 = 1. The equation with roots 2 and 1 is x^2 - (2+1)x + (2*1) = 0, which is x^2 - 3x + 2 = 0.

Multiple choice

If $\alpha &amp; \beta$ are roots of quadratic equation ${ x }^{ 2 }+13x+8=0$ then the value of ${ \alpha }^{ 4 }+{ \beta }^{ 4 }=$
  1. $23281$
  2. $23218$
  3. $23128$
  4. $23182$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For x^2 + 13x + 8 = 0, alpha + beta = -13 and alpha * beta = 8. We need alpha^4 + beta^4. First, alpha^2 + beta^2 = (alpha + beta)^2 - 2(alpha * beta) = (-13)^2 - 2(8) = 169 - 16 = 153. Then, alpha^4 + beta^4 = (alpha^2 + beta^2)^2 - 2(alpha * beta)^2 = 153^2 - 2(8^2) = 23409 - 128 = 23281.

Multiple choice

If ${\text{a}}\;\;$ and ${\text{b}}\;$ are the roots of equation ${{\text{x}}^{\text{2}}}{\text{ + px + 1 = 0}}\;{\text{and}}\;{\text{c,}}$ d are the roots of equation $\;\;\;{{\text{x}}^{\text{2}}}{\text{ + qx + 1 = 0}}$, then the value of ${\text{E = }}\left( {{\text{a - c}}} \right)\left( {{\text{b - c}}} \right)\left( {{\text{a + d}}} \right)\left( {{\text{b + d}}} \right)$ is
  1. ${p^2} - {q^2}$
  2. ${q^2} - {p^2}$
  3. ${q^2} + {p^2}$

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

If the roots of the equation $a{x^2} + bx + c = 0$ are of the form $\displaystyle {{K + 1} \over K}$ and $\displaystyle{{K + 2} \over {K + 1}}$ then sum of roots is equal to;
  1. ${(a + b + c)^2}$
  2. ${(a + b - c)^2}$
  3. ${(a - b + c)^2}$

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

The sum of roots is -b/a. Given roots are (K+1)/K and (K+2)/(K+1). Sum = 1 + 1/K + 1 + 1/(K+1) = 2 + 1/K + 1/(K+1). This does not simplify to a standard form related to a, b, c without further constraints.