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 $-5$ is a root of the quadratic $2{x^2} + px - 15 = 0$ and the quadratic equation $p({x^2} + x) + k = 0\,\,$ has equal root , find the value of $k.$
  1. $\dfrac {7}{4}$
  2. $\dfrac {7}{2}$
  3. $-\dfrac {7}{4}$

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

If -5 is a root of 2x^2 + px - 15 = 0, then 2(25) - 5p - 15 = 0, so 50 - 15 = 5p, 35 = 5p, p = 7. The second equation is 7(x^2 + x) + k = 0, or 7x^2 + 7x + k = 0. For equal roots, discriminant D = b^2 - 4ac = 0. So 7^2 - 4(7)(k) = 0, 49 - 28k = 0, k = 49/28 = 7/4.

Multiple choice

For which value of $k$ will the equations ${x}^{2}-kx-21=0$ and ${x}^{2}-3kx+35=0$ have one common root.
  1. $\pm 1$
  2. $\pm 2$
  3. $\pm 4$

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

If the equations have a common root 'a', then a^2 - ka - 21 = 0 and a^2 - 3ka + 35 = 0. Subtracting the two equations gives 2ka - 56 = 0, so ka = 28. Substituting ka = 28 into the first equation gives a^2 - 28 - 21 = 0, so a^2 = 49, meaning a = 7 or -7. If a = 7, then 7k = 28, so k = 4. If a = -7, then -7k = 28, so k = -4.

Multiple choice

If the quadratic equations $3x^{2} + ax + 1 = 0$ and $2x^{2} + bx + 1 = 0$ have a common root, then the variable point $(a, b)$ moves on

  1. A circle

  2. A parabola

  3. An ellipse

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

If alpha is the common root, 3alpha^2 + a*alpha + 1 = 0 and 2alpha^2 + b*alpha + 1 = 0. Subtracting: alpha^2 + (a-b)alpha = 0. So alpha = 0 (impossible as 1=0) or alpha = b-a. Substituting alpha = b-a into the first: 3(b-a)^2 + a(b-a) + 1 = 0. This is a quadratic in a and b representing a hyperbola.

Multiple choice

The root of the equation $|x^{2}-x-6|=x+2$ are given by:
  1. $-2,2,4$
  2. $0,1,4$
  3. $-2,1,4$
  4. $0,2,4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solve |x^2 - x - 6| = x + 2. Case 1: x^2 - x - 6 = x + 2 => x^2 - 2x - 8 = 0 => (x-4)(x+2) = 0 => x = 4, -2. Case 2: x^2 - x - 6 = -(x + 2) => x^2 - 4 = 0 => x = 2, -2. Combining gives -2, 2, 4.

Multiple choice

Find the values of $\lambda$ so that the equations $x^2-\lambda x -21=0$ and $x^2-3\lambda x +35=0$ may have one common root.
  1. $\pm 1$
  2. $\pm 2$
  3. $\pm 3$
  4. $\pm 4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For common root alpha, alpha^2 - lambda*alpha - 21 = 0 and alpha^2 - 3*lambda*alpha + 35 = 0. Subtracting: 2*lambda*alpha - 56 = 0 => lambda*alpha = 28. Substituting into first: alpha^2 - 28 - 21 = 0 => alpha^2 = 49 => alpha = +/- 7. If alpha = 7, lambda = 4. If alpha = -7, lambda = -4.

Multiple choice

If $\alpha$, $\beta$ are the roots of the equation $ { x }^{ 2 }+x+2=0$, then what is $\dfrac { { \alpha }^{ 10 }+{ \beta }^{ 10 } }{ { \alpha }^{ -10 }+{ \beta }^{ -10 } } $ equal to?
  1. $4096$
  2. $2048$
  3. $1024$
  4. $512$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given x^2 + x + 2 = 0, roots are alpha, beta. alpha + beta = -1, alpha * beta = 2. Expression = (alpha^10 + beta^10) / ((alpha^10 + beta^10) / (alpha^10 * beta^10)) = (alpha * beta)^10 = 2^10 = 1024.

Multiple choice

If the equations $ 3x^2 - 2x +p =0 $ and $ 6x^2 -17x +12 = 0 $ have a common root , then the value of p is
  1. $\dfrac{15}{3} , \dfrac{8}{3}$
  2. $\dfrac{-15}{4} , \dfrac{-8}{3}$
  3. $\dfrac{15}{7} , \dfrac{8}{3}$
  4. $\dfrac{-15}{4} , \dfrac{8}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Roots of 6x^2 - 17x + 12 = 0 are (17 +/- sqrt(289 - 288))/12 = (17 +/- 1)/12. Roots are 18/12 = 3/2 and 16/12 = 4/3. If 3/2 is a common root: 3(9/4) - 2(3/2) + p = 0 => 27/4 - 3 + p = 0 => p = -15/4. If 4/3 is a common root: 3(16/9) - 2(4/3) + p = 0 => 16/3 - 8/3 + p = 0 => p = -8/3.

Multiple choice

$If\quad the\quad equations\quad { x }^{ 2 }+ax+b=0\quad and\quad { x }^{ 2 }+px-q=0\quad have\quad a\quad coomon\quad root\ then\quad that\quad root\quad is$
  1. $\frac { b+q }{ p-a }  $
  2. $\frac { b-q }{ p-q } or\frac { a-b }{ p-q } $
  3. $\frac { b+q }{ p+a } or\frac { aq-bp }{ p-q } $
  4. $\frac { b-q }{ p-a } or\frac { aq-bp }{ b-q } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let alpha be the common root. alpha^2 + a*alpha + b = 0 and alpha^2 + p*alpha - q = 0. Subtracting: (a-p)*alpha + (b+q) = 0. alpha = (b+q) / (p-a).