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

The number of integral values of $a$ for which the quadratic equation $\left ( a+2 \right )x^{2}+2\left ( a+1 \right )x+a=0$ will have integer roots are
  1. $3$
  2. $2$
  3. $5$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the equation (a+2)x^2 + 2(a+1)x + a = 0, the roots are x = [-2(a+1) +/- sqrt(4(a+1)^2 - 4a(a+2))] / 2(a+2). Simplifying the discriminant gives sqrt(4a^2 + 8a + 4 - 4a^2 - 8a) = sqrt(4) = 2. The roots are x = [-2a - 2 +/- 2] / 2(a+2), which are -1 and -a/(a+2). For -a/(a+2) to be an integer, a+2 must divide -a. Since -a = -(a+2) + 2, a+2 must divide 2. Possible values for a+2 are 1, -1, 2, -2, leading to a = -1, -3, 0, -4. Checking these, all yield integer roots.

Multiple choice

If $\alpha , \beta $ are roots of equation $x^{2}-4x-3=0$ and $s_{n}= \alpha ^{n} + \beta ^{n},n \in N$, then the value of $\displaystyle \frac{s_{7}-4s_{6}}{s_{5}}$ is
  1. $3$
  2. $4$
  3. $5$
  4. $7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since alpha and beta are roots of x^2 - 4x - 3 = 0, alpha^2 = 4*alpha + 3 and beta^2 = 4*beta + 3. Multiplying by alpha^(n-2) and beta^(n-2) and adding gives s_n = 4*s_(n-1) + 3*s_(n-2). Thus, s_7 - 4*s_6 = 3*s_5. The ratio is 3.

Multiple choice

If $\alpha ,\beta ,\gamma $ are the roots of the cubic ${ x }^{ 3 }+qx+r=0$, then the equation whose roots are $ { \left( \alpha -\beta \right) }^{ 2 },{ \left( \beta -\gamma \right) }^{ 2 },{ \left( \gamma -\alpha \right) }^{ 2 }$, is:
  1. $ { y }^{ 3 }+6q{ y }^{ 2 }+9{ q }^{ 2 }y+\left( { 4q }^{ 3 }+{ 27r }^{ 2 } \right) =0$
  2. $ { y }^{ 3 }+6q{ y }^{ 2 }+9{ q }^{ 2 }y+\left( { 4r }^{ 3 }-{ 27r }^{ 2 } \right) =0$
  3. $ { y }^{ 3 }+6q{ y }^{ 2 }+9{ q }^{ 2 }y+\left( { 4r }^{ 3 }+{ 27q }^{ 2 } \right) =0$
  4. $ { y }^{ 3 }+6q{ y }^{ 2 }+9{ q }^{ 2 }y+\left( { 4q }^{ 3 }-{ 27r }^{ 2 } \right) =0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a standard problem in the theory of equations. For the cubic x^3 + qx + r = 0, the transformation to find the equation with roots (alpha-beta)^2 etc., leads to the result y^3 + 6qy^2 + 9q^2y + (4q^3 + 27r^2) = 0.

Multiple choice

If the quadratic equation $a{ x }^{ 2 } + bx + c = 0; a>0$ has real roots of opposite sign in the interval $\left( -2, 2 \right)$, then comment on the value of the following expression $1 + \displaystyle\frac { c }{ 4a } - \left| \displaystyle\frac { b }{ 2a } \right| $.
  1. $(0,\infty)$
  2. $(-\infty,0)$
  3. $(1-\sqrt{3},1+\sqrt{3})$
  4. $(-\infty,1-\sqrt{3})\cup (1+\sqrt{3},\infty)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given roots are real and of opposite sign, c/a < 0. The expression is 1 + c/4a - |b/2a|. Since the roots are in (-2, 2), the sum of roots -b/a is in (-4, 4), so |b/2a| < 2. Also, c/a is the product of roots, so c/a > -4. The expression simplifies to values in (0, infinity).

Multiple choice

If $\displaystyle \alpha, \beta$ are the roots of the quadratic equation $\displaystyle 4x^2 - 16x + k = 0, k\in R$ such that $\displaystyle 1 < \alpha < 2$ and $\displaystyle 2 < \beta < 3$, then the number of integral values of $k$ is
  1. $5$
  2. $6$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let f(x) = 4x^2 - 16x + k. The roots alpha and beta satisfy 1 < alpha < 2 and 2 < beta < 3. This implies f(1) > 0, f(2) < 0, and f(3) > 0. f(1) = 4 - 16 + k = k - 12 > 0 => k > 12. f(2) = 16 - 32 + k = k - 16 < 0 => k < 16. f(3) = 36 - 48 + k = k - 12 > 0 => k > 12. The range is 12 < k < 16. The integers are 13, 14, 15. There are 3 values.

Multiple choice

If exactly one root of the quadratic equation $\displaystyle x^2 - (a + 1) x + 2a = 0$ lies in the interval $(0, 3)$, then the set of values $a$ is given by
  1. $\displaystyle (-\infty, 0) \cup (6, \infty)$
  2. $\displaystyle (-\infty, 0] \cup (6, \infty)$
  3. $\displaystyle (-\infty, 0] \cup [6, \infty)$
  4. $\displaystyle (0, 6)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let f(x) = x^2 - (a+1)x + 2a = (x-a)(x-2). Roots are a and 2. For exactly one root in (0, 3), we check the boundary conditions. If a=2, both roots are 2 (not 'exactly one'). If a <= 0, root 2 is in (0, 3) and root a is <= 0 (outside). If a > 6, root 2 is in (0, 3) and root a > 6 (outside). If a=6, roots are 6 and 2 (only 2 is in range).

Multiple choice

If $ \alpha, \beta $ are the roots of the equation $ 8x^2-3x+27=0 $, then the value of $\left( \dfrac { \alpha ^{ 2 } }{ \beta } \right) ^{ \frac { 1 }{ 3 } }+\left( \dfrac { \beta ^{ 2 } }{ \alpha } \right) ^{ \frac { 1 }{ 3 } }$ is
  1. $ \dfrac{1}{3} $
  2. $ \dfrac{1}{4} $
  3. $ \dfrac{7}{2} $
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 8x^2 - 3x + 27 = 0, alpha + beta = 3/8 and alpha * beta = 27/8. The expression is (alpha^2/beta)^(1/3) + (beta^2/alpha)^(1/3) = (alpha + beta) / (alpha * beta)^(1/3). This simplifies to (3/8) / (27/8)^(1/3) = (3/8) / (3/2) = 3/8 * 2/3 = 1/4.

Multiple choice

Find all the roots of the equation $\displaystyle 4x^{4}-24x^{3}+57x^{2}+18x-45= 0$, if one of them is $\displaystyle 3+i\sqrt{6}.$
  1. $\displaystyle 3\pm i\sqrt{6};\> \pm \dfrac{\sqrt{3}}{2}$
  2. $\displaystyle \pm3+ i\sqrt{6};\> \pm \dfrac{\sqrt{3}}{2} $
  3. $\displaystyle 3\pm i\sqrt{6};\> \dfrac{\sqrt{3}}{2}$
  4. $\displaystyle \pm3+ i\sqrt{6};\> - \dfrac{\sqrt{3}}{2} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since coefficients are real, if 3 + i*sqrt(6) is a root, 3 - i*sqrt(6) is also a root. The quadratic factor is (x - (3 + i*sqrt(6)))(x - (3 - i*sqrt(6))) = (x - 3)^2 + 6 = x^2 - 6x + 15. Dividing 4x^4 - 24x^3 + 57x^2 + 18x - 45 by x^2 - 6x + 15 gives 4x^2 - 3 = 0. Roots are x^2 = 3/4, x = +/- sqrt(3)/2.