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 $\alpha , \beta$ be the roots of $x ^ { 2 } + x + 1 = 0$ . the equation whose roots are $\alpha ^ { 19 }$ and $\beta ^ { 16 }$ is
  1. $x ^ { 2 } + x + 1 = 0$
  2. $x ^ { 2 } - x + 1 = 0$
  3. $x ^ { 2 } + x - 1 = 0$

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

Roots of x^2+x+1=0 are omega and omega^2. omega^3=1. alpha^19 = omega^19 = omega. beta^16 = (omega^2)^16 = omega^32 = omega^2. The roots are omega and omega^2, so the equation is the same.

Multiple choice

If the roots of the equation $x ^ { 2 } - 2 a x + a ^ { 2 } + a - 3 = 0$ are real and greater than $3 ,$ then
  1. $a \in \phi$
  2. $a > 3$
  3. $a < 2$ or $a > 3$
  4. $a \leq 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For real roots, D >= 0: (2a)^2 - 4(a^2 + a - 3) >= 0 => 4a^2 - 4a^2 - 4a + 12 >= 0 => -4a + 12 >= 0 => a <= 3. For roots > 3, f(3) > 0 and vertex > 3. f(3) = 9 - 6a + a^2 + a - 3 = a^2 - 5a + 6 > 0 => (a-2)(a-3) > 0 => a < 2 or a > 3. Vertex = 2a/2 = a > 3. Intersection of a <= 3, (a < 2 or a > 3), and a > 3 is empty.

Multiple choice

If the roots of the equation $b x ^ { 2 } + c x + a = 0$ be imaginary, then for all real values of $x ,$ the expression $3 b ^ { 2 } x ^ { 2 } + 6 b c x + 2 c ^ { 2 }$ is -
  1. Greater than $- 4 a b$
  2. Less than $- 4 a b$
  3. Greater than $4\mathrm { ab }$
  4. Less than $4\mathrm { ab }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If bx^2 + cx + a = 0 has imaginary roots, the discriminant c^2 - 4ab < 0, so c^2 < 4ab. The expression 3b^2x^2 + 6bcx + 2c^2 can be written as 3(bx + c)^2 - c^2. Since c^2 < 4ab, -c^2 > -4ab. Thus, the expression is greater than -4ab.

Multiple choice

$p,:q,:r,:$ and $:s$ are integers. If the $AM$ of the roots of the quadratic equation $\displaystyle x^{2}-px+q^{2}=0$ and $GM$ of the roots of the quadratic equation $\displaystyle x^{2}-rx+s^{2}=0$ are equal then
  1. $q$ is an odd integer
  2. $r$ is an even integer
  3. $p$ is an even integer
  4. $s$ is an odd integer
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

AM of roots of x^2 - px + q^2 = 0 is p/2. GM of roots of x^2 - rx + s^2 = 0 is sqrt(s^2) = |s|. So p/2 = |s|, p = 2|s|. Since s is an integer, p must be an even integer.

Multiple choice

If $a> 1$, roots of the equation $\displaystyle (1-a)x^{2}+3ax-1=0$ are

  1. one positive and one negative

  2. both negative

  3. both positive

  4. both nonreal complex
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation is (1-a)x^2 + 3ax - 1 = 0. Since a > 1, the coefficient of x^2 is negative. The product of roots is c/a = -1 / (1-a) = 1 / (a-1), which is positive. The sum of roots is -b/a = -3a / (1-a) = 3a / (a-1), which is also positive. Since both sum and product are positive, both roots must be positive.

Multiple choice

If the sum of the roots of the quadratic equation $\displaystyle ax^{2}+bx+c=0$ is equal to the sum of the squares of their reciprocals then $\displaystyle \frac{b^{2}}{ac}+\frac{bc}{a^{2}}$ is equal to
  1. $2$
  2. $-2$
  3. $1$
  4. $-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sum of roots = -b/a. Sum of squares of reciprocals = (1/r1)^2 + (1/r2)^2 = (r1^2 + r2^2) / (r1*r2)^2 = ((r1+r2)^2 - 2r1r2) / (r1r2)^2. Equating -b/a = (b^2/a^2 - 2c/a) / (c^2/a^2) leads to -b/a = (b^2 - 2ac) / c^2. Rearranging yields b^2/ac + bc/a^2 = 2.

Multiple choice

If the equation $x ^ { 2 } + 5 b x + 8 c = 0 ,$ does not have two distinct real roots, then minimum value of $5 b +8c$ is
  1. $-1$
  2. $-2$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the equation x^2 + 5bx + 8c = 0 to not have two distinct real roots, the discriminant D must be <= 0. D = (5b)^2 - 4(1)(8c) = 25b^2 - 32c <= 0. This implies 32c >= 25b^2. We want the minimum value of 5b + 8c. Since 8c >= (25/4)b^2, we minimize f(b) = 5b + (25/4)b^2. Derivative f'(b) = 5 + (25/2)b = 0, so b = -10/25 = -0.4. Min value = 5(-0.4) + (25/4)(-0.4)^2 = -2 + (25/4)(0.16) = -2 + 1 = -1.

Multiple choice

If the equation ${ x }^{ 2 }-px+q=0$ and ${ x }^{ 2 }-ax+b=0$ have a common root and the other roots of the second equation is the reciprocal of the other root of the first, then ${ \left( q-b \right) }^{ 2 }=$
  1. $aq{ \left( p-b \right) }^{ 2 }$
  2. $bq{ \left( p-a \right) }^{ 2 }$
  3. $bq{ \left( p-b \right) }^{ 2 }$

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

Let the common root be alpha. alpha^2 - p*alpha + q = 0 and alpha^2 - a*alpha + b = 0. Subtracting: (a-p)*alpha + (q-b) = 0, so alpha = (b-q)/(a-p). Substituting back and using the reciprocal root property, the identity (q-b)^2 = bq(p-a)^2 holds.