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

How many roots can be common between two Quadratic Equations?
  1. Only $1$
  2. More than $2$
  3. Not more than $2$

  4. No roots can be common
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A quadratic equation has at most 2 roots. Two distinct quadratic equations can share at most 2 roots (if they are identical) or 1 root (if they are not). Thus, the maximum number of common roots is 2.

Multiple choice

Find the value of 'k' for which the following set of quadratic equation has exactly one common root ${x^{2 }}-kx + 10 =0 $ and ${x^2} + kx - 18 = 0.$
  1. $ \pm 3$
  2. $ \pm 7$
  3. $ \pm 9$
  4. $ \pm 11$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a common root 'a', we have a^2 - ka + 10 = 0 and a^2 + ka - 18 = 0. Subtracting the equations gives 2ka = 28, so ka = 14. Substituting a = 14/k into the first equation gives (14/k)^2 - 14 + 10 = 0, so 196/k^2 = 4, which means k^2 = 49, so k = +/- 7.

Multiple choice

lf the equations $ax^{2}+2bx+3c=0$ and $3x^{2}+8x+15=0$ have a common root, where $a,\ b,\ c$ are the length of the sides of a $\Delta \mathrm{A}BC$, then $\sin^{2}\mathrm{A}+\sin^{2}B+\sin^{2}C$ is equal to
  1. $1$
  2. $\displaystyle \frac{3}{2}$
  3. $\sqrt{2}$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation 3x^2 + 8x + 15 = 0 has complex roots. For the equations to have a common root, the coefficients must be proportional. Comparing ax^2 + 2bx + 3c = 0 and 3x^2 + 8x + 15 = 0, we get a/3 = 2b/8 = 3c/15, so a/3 = b/4 = c/5. Since a, b, c are sides of a triangle, they can be 3, 4, 5. This is a right triangle with hypotenuse 5. Thus, sin^2 A + sin^2 B + sin^2 C = (3/5)^2 + (4/5)^2 + (5/5)^2 = 9/25 + 16/25 + 1 = 1 + 1 = 2.

Multiple choice

If the equation $px^{2}+qx+r=0$ and $rx^{2}+qx+p=0$, where $p\neq r$, have a negative common root, then value of $p-q$ is
  1. $r$
  2. $2r$
  3. $-r$

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

Let alpha be the common root. p(alpha)^2 + q(alpha) + r = 0 and r(alpha)^2 + q(alpha) + p = 0. Subtracting: (p-r)(alpha)^2 + (r-p) = 0. (p-r)(alpha^2 - 1) = 0. Since p != r, alpha^2 = 1. Since the root is negative, alpha = -1. Substitute alpha = -1 into the first equation: p - q + r = 0. Thus, p - q = -r.

Multiple choice

For $\sqrt { 4+\sqrt { 83 } } $, the correct option is .....................

  1. Does not exist as quadratic surd

  2. Does not exist as real numbers
  3. $2+\sqrt { 83 } $
  4. $\sqrt { 83 }-2 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A quadratic surd is of the form sqrt(a + sqrt(b)). For it to be simplified to sqrt(x) + sqrt(y), (a^2 - b) must be a perfect square. Here, 4^2 - 83 = 16 - 83 = -67, which is not a perfect square. Thus, it cannot be simplified into a simple surd form.

Multiple choice

If $HCF$ & $LCM$ of two quadratic expression are $(x-5)$ and $x^3-19x-30$ then find the two expressions.
  1. $p(x) = (x-5) (x+2) and\ q(x) = (x-5)(x+3)$
  2. $p(x) = (x-2) (x-8)\ and\  q(x) = (x-1)(x-3)$
  3. $p(x) = (x+1) (x-2)\ and\  q(x) = (x-1)(x-3)$
  4. $p(x) = (x-1) (x-2)\ and\  q(x) = (x-1)(x-8)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The HCF is (x-5). The LCM is x^3 - 19x - 30. Factoring the LCM: (x-5)(x^2+5x+6) = (x-5)(x+2)(x+3). The two expressions must be formed by distributing these factors such that each contains the HCF (x-5). Option A gives (x-5)(x+2) and (x-5)(x+3), which share (x-5) as HCF and their LCM is (x-5)(x+2)(x+3) = x^3 - 19x - 30.

Multiple choice

The roots of both the equations $\sin ^{ 2 }{ x } +p\sin { x } +q=0\quad $ and $\cos ^{ 2 }{ x } +r\cos { x } +s=0$ are $\alpha,\beta$. $ Then \ the \ value \ of \sin { \left( \alpha +\beta \right) }=$
  1. $\cfrac { 2qs }{ { q }^{ 2 }+{ s }^{ 2 } } $
  2. $\cfrac { 2ps }{ { p }^{ 2 }+{ s }^{ 2 } } $
  3. $\cfrac { 2pr }{ { p }^{ 2 }+{ r }^{ 2 } } $
  4. $\cfrac { 2qr }{ { q }^{ 2 }+{ r }^{ 2 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For sin^2(x) + p*sin(x) + q = 0, roots are sin(alpha) and sin(beta). So sin(alpha) + sin(beta) = -p and sin(alpha)*sin(beta) = q. For cos^2(x) + r*cos(x) + s = 0, roots are cos(alpha) and cos(beta). So cos(alpha) + cos(beta) = -r and cos(alpha)*cos(beta) = s. Using sin(alpha+beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta), this is complex. However, standard identity manipulation for these specific quadratic forms leads to 2pr / (p^2 + r^2).