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

Let p, q, and r be real numbers $(p\neq q, r\neq 0)$, such that the roots of the equation $\displaystyle\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to.
  1. $p^2+q^2+r^2$
  2. $p^2+q^2$
  3. $2(p^2+q^2)$
  4. $\displaystyle\frac{p^2+q^2}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is (x+q + x+p) / (x+p)(x+q) = 1/r, which simplifies to r(2x + p + q) = x^2 + (p+q)x + pq. Since roots are equal in magnitude but opposite in sign, their sum is 0. The sum of roots is -(p+q-2r) = 0, so p+q = 2r. The product of roots is pq - rq = -k^2. Solving for the sum of squares of roots (k^2 + (-k)^2 = 2k^2) leads to p^2 + q^2.

Multiple choice

Let p(x) be a quadratic polynomial such that $p(0)=1$. If p(x) leaves remainder $4$ when divided by $x-1$ and it leaves remainder $6$ when divided by $x+1$; then which one is correct?
  1. $p(2)=11$
  2. $p(-2)=19$
  3. $p(-2)=11$
  4. $p(2)=19$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let p(x) = ax^2 + bx + c. p(0)=1 implies c=1. p(1)=4 implies a+b+1=4 => a+b=3. p(-1)=6 implies a-b+1=6 => a-b=5. Adding: 2a=8, a=4. Subtracting: 2b=-2, b=-1. So p(x) = 4x^2 - x + 1. p(-2) = 4(4) - (-2) + 1 = 16 + 2 + 1 = 19.

Multiple choice

If $2 + 3i$ is one of the roots of the equation $2x^3 - 9x^2 + kx - 13 = 0,\,\, k \in R$, then the real root of this equation :
  1. exist and is equal to $\displaystyle \frac{1}{2}$

  2. does not exist

  3. exist and is equal to 1
  4. exist and is equal to $- \displaystyle \frac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since coefficients are real, if 2 + 3i is a root, 2 - 3i must also be a root. The sum of roots is (2 + 3i) + (2 - 3i) + r = 9/2, so 4 + r = 4.5, which means r = 0.5 or 1/2.

Multiple choice

If the two roots of the equation, $(a - 1)( x^4 + x^2 + 1) + (a + 1) (x^2 + x + 1)^2 = 0$ are real and distinct, then the set of all values of $a$ is:
  1. $\displaystyle \left ( 0, \frac{1}{2} \right )$
  2. $\displaystyle \left (- \frac{1}{2}, 0 \right ) \cup \left ( 0, \frac{1}{2} \right )$
  3. $(- \infty, - 2) \cup (2, \infty)$
  4. $\displaystyle \left ( -\frac{1}{2}, 0 \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation can be rewritten by dividing by (x^2+x+1)^2. Let y = (x^2+x+1)/(x^2-x+1) or similar substitutions. The condition for real and distinct roots leads to the interval (-1/2, 0) union (0, 1/2).

Multiple choice

If $\alpha$ and $\beta$ are roots of the equation, $x^2-4\sqrt{2} kx+2e^{4\ln k} - 1=0$ for some $k$, and $\alpha^2+\beta^2=66$, then $\alpha^3+\beta^3$ is equal to
  1. $-32 \sqrt{2}$
  2. $280 \sqrt{2}$
  3. $-280 \sqrt{2}$
  4. $248 \sqrt{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is x^2 - 4*sqrt(2)*k*x + 2*k^4 - 1 = 0. Sum of roots (a+b) = 4*sqrt(2)*k and product (ab) = 2*k^4 - 1. Given a^2+b^2 = (a+b)^2 - 2ab = 66, we have 32*k^2 - 2(2*k^4 - 1) = 66, which simplifies to -4*k^4 + 32*k^2 - 64 = 0, or k^4 - 8*k^2 + 16 = 0. This is (k^2 - 4)^2 = 0, so k^2 = 4. Then a+b = 8*sqrt(2) and ab = 2(16)-1 = 31. a^3+b^3 = (a+b)((a+b)^2 - 3ab) = 8*sqrt(2)(128 - 93) = 8*sqrt(2)(35) = 280*sqrt(2).

Multiple choice

If equations $ax^2+bx+c=0, (a, b, c \in R, a\neq 0)$ and $2x^2+3x+4=0$ have a common root, then $a : b : c$ equals :
  1. $1 : 2 : 3$
  2. $2 : 3 : 4$
  3. $4 : 3 : 2$
  4. $3 : 2 : 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If ax^2 + bx + c = 0 and 2x^2 + 3x + 4 = 0 have a common root, the coefficients must be proportional if the roots are identical, or satisfy the resultant condition. Comparing the equations, if they share the same roots, a/2 = b/3 = c/4, so a:b:c = 2:3:4.

Multiple choice

If $\dfrac {1}{\sqrt{\alpha}}$ and $\dfrac {1}{\sqrt{\beta}}$ are the roots of the equation, $ax^2+bx+1=0$ $(a\neq 0, a, b \in R)$, then the equation, $x(x+b^3)+(a^3-3abx)=0$ has roots
  1. $\alpha^{{3}/{2}}$ and $\beta^{{3}/{2}}$
  2. $\alpha \beta^{{1}/{2}}$ and $\alpha^{{1}/{2}}\beta$
  3. $\sqrt {\alpha \beta}$ and $\alpha \beta$
  4. $\alpha^{-{3}/{2}}$ and $\beta^{- {3}/{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given 1/sqrt(alpha) and 1/sqrt(beta) are roots of ax^2 + bx + 1 = 0. This implies alpha and beta are roots of a transformed equation. Through algebraic manipulation of the coefficients and roots, the roots of the second equation are alpha^(3/2) and beta^(3/2).

Multiple choice

Let $ \alpha$ and $ \beta$ be the roots of equation $ px^{2}+qx+r=0,p\neq 0$. If $ p,q,r$ are in A.P. and $ \displaystyle \frac{1}{\alpha }+\displaystyle \frac{1}{\beta }=4$, then the value of $ |\alpha -\beta |$ is
  1. $ \displaystyle \frac{\sqrt{61}}{9}$
  2. $ \displaystyle \frac{2\sqrt{17}}{9}$
  3. $ \displaystyle \frac{\sqrt{34}}{9}$
  4. $ \displaystyle \frac{2\sqrt{13}}{9}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Roots alpha, beta satisfy alpha+beta = -q/p and alpha*beta = r/p. 1/alpha + 1/beta = (alpha+beta)/(alpha*beta) = (-q/p) / (r/p) = -q/r = 4. So q = -4r. Since p, q, r are in AP, 2q = p+r. Substitute q: -8r = p+r => p = -9r. Equation: -9rx^2 - 4rx + r = 0. Divide by -r: 9x^2 + 4x - 1 = 0. |alpha-beta| = sqrt((alpha+beta)^2 - 4*alpha*beta) = sqrt((-4/9)^2 - 4*(-1/9)) = sqrt(16/81 + 36/81) = sqrt(52/81) = sqrt(4*13)/9 = 2*sqrt(13)/9.

Multiple choice

lf $\alpha$ and $\beta$ are the roots of the equation $\mathrm{x}^{2}-\mathrm{x}+1 =0$, then $\alpha^{2009}+\beta^{2009}=$
  1. $-1$
  2. $1$
  3. $2$
  4. $-2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The roots of x^2 - x + 1 = 0 are -omega and -omega^2 (where omega is the cube root of unity). alpha^3 = -1, so alpha^6 = 1. alpha^2009 = (alpha^6)^334 * alpha^5 = alpha^5 = -omega^5 = -omega^2. Similarly, beta^2009 = -omega. The sum is -(omega + omega^2) = -(-1) = 1.

Multiple choice

The common roots of the equation $z^{3} + 2z^{2} + 2z + 1 = 0, z^{2014} + z^{2015} + 1 = 0$ are
  1. $\omega, \omega^{2}$
  2. $1, \omega, \omega^{2}$
  3. $-1, \omega, \omega^{2}$
  4. $-\omega, -\omega^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots of z^3 + 2z^2 + 2z + 1 = 0 are z = -1 and the roots of z^2 + z + 1 = 0 (which are omega and omega^2). The roots of z^2014 + z^2015 + 1 = 0 are omega and omega^2 because omega^2 + omega + 1 = 0. Thus, the common roots are omega and omega^2.

Multiple choice

If p and q are distinct prime numbers and if the equation $x^2-px+q=0$ has positive integers as its roots then the roots of the equation are

  1. 1, -1

  2. 2, 3

  3. 1, 2

  4. 3, 1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the quadratic equation x^2 - px + q = 0, the sum of roots is p and the product is q. Since p and q are primes and roots are positive integers, let the roots be r1 and r2. r1 * r2 = q implies one root must be 1 and the other q (since q is prime). Then r1 + r2 = 1 + q = p. For both p and q to be prime, the only solution is q = 2, which makes p = 3. Thus the roots are 1 and 2.