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

lf $\alpha,\ \beta$ are the roots of the equation $x^{2}+x+1=0$ and $S_k={\alpha}^k+{\beta}^k ; k=1,2,3,4 $ , then $\left| \begin{matrix} 3 & 1+{ S }{ 1 } & 1+{ S }{ 2 } \ 1+{ S }{ 1 } & 1+{ S }{ 2 } & 1+{ S }{ 3 } \ 1+{ S }{ 2 } & 1+{ S }{ 3 } & 1+{ S }{ 4 } \end{matrix} \right| =\ $
  1. $27$
  2. $-27$
  3. $-3$
  4. $9$
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. S_k = omega^k + omega^(2k). S_1 = -1, S_2 = -1, S_3 = 2, S_4 = -1. The determinant evaluates to -27.

Multiple choice

The quadratic equation whose roots are $l,m $ where $l=\displaystyle \lim_{\theta \rightarrow 0}\ \displaystyle \frac{3\sin\theta-4\sin^{2}\theta}{\theta}$, $m=\displaystyle \lim_{\theta \rightarrow 0}\ \frac{2\tan\theta }{\theta}$is
  1. $\mathrm{x}^{2}-5\mathrm{x}+6=0$
  2. $\mathrm{x}^{2}+5\mathrm{x}+6=0$
  3. $\mathrm{x}^{2}-\mathrm{x}+6=0$
  4. $\mathrm{x}^{2}-\mathrm{x}-6=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The limit l is calculated as lim(theta->0) (3sin(theta)/theta - 4sin^2(theta)/theta) = 3 - 0 = 3. The limit m is calculated as lim(theta->0) (2tan(theta)/theta) = 2. The quadratic equation with roots 3 and 2 is (x-3)(x-2) = x^2 - 5x + 6 = 0.

Multiple choice

If $x_1$ and $x_2$ are the roots of the equation $e^2.x^{\ln x}=x^3$ with $x_1>x_2$, then what is the relation between the roots?
  1. $x_1=2x_2$
  2. $x_1=(x_2)^2$
  3. $2x_1=(x_2)^2$
  4. $(x_1)^2=(x_2)^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Taking the natural log of both sides: ln(e^2 * x^ln x) = ln(x^3) becomes 2 + (ln x)^2 = 3 ln x. Let u = ln x, then u^2 - 3u + 2 = 0. Roots are u = 1, 2. Thus ln x1 = 2 (x1 = e^2) and ln x2 = 1 (x2 = e). Therefore, x1 = (x2)^2.

Multiple choice

Let $a$, $b$ and $c$ be the sides of a $ \Delta ABC$. If $a^2, b^2 :and :c^2$ are the roots of the equation $\left (x^3 - Px^2 + Qx - R = 0\right )$, where P, Q and R are constants then the value of $ \left (\dfrac{cosA}{a}+\dfrac{cosB}{b}+\dfrac{cosC}{c}\right )$ is equal to
  1. $\dfrac{P}{2\sqrt{R}}$
  2. $\dfrac{P}{\sqrt{2R}}$
  3. $\dfrac{P\sqrt{2}}{R}$

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

By Vieta's formulas for the given cubic equation, the sum of the roots is a^2 + b^2 + c^2 = P, and the product of the roots is a^2 * b^2 * c^2 = R, which gives abc = sqrt(R). Using the law of cosines, we can write cosA/a = (b^2 + c^2 - a^2) / (2abc). Summing this expression for all three angles simplifies to (a^2 + b^2 + c^2) / (2abc), which equals P / (2*sqrt(R)).

Multiple choice

If $\alpha,\beta,\gamma$ are the roots of the equation $x^3-x-1=0$, then the value of $\sum \left(\dfrac {1+\alpha}{1-\alpha}\right)$ is-
  1. $-3$
  2. $-5$
  3. $-7$

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

Given x^3 - x - 1 = 0, the roots are alpha, beta, gamma. We need the sum of (1+r)/(1-r) for each root. Let y = (1+x)/(1-x), then x = (y-1)/(y+1). Substituting this into the original equation: ((y-1)/(y+1))^3 - (y-1)/(y+1) - 1 = 0. Expanding this leads to a cubic in y, and the sum of roots is -coefficient of y^2 / coefficient of y^3, which results in -7.

Multiple choice

If positive numbers $a,b,c$ be in H.P, then the equation $x^2-kx+2b^{101}-a^{101}-c^{101}=0$ $(k\in R)$ has

  1. Both roots positive

  2. Both roots negative

  3. One positive and one negative root

  4. Both roots imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since a, b, c are in H.P., 2/b = 1/a + 1/c, which implies b = 2ac/(a+c). The constant term of the quadratic is 2b^101 - a^101 - c^101. By AM-GM inequality, b^101 < (a^101 + c^101)/2, so 2b^101 < a^101 + c^101, making the constant term negative. A quadratic with a negative constant term always has roots of opposite signs.

Multiple choice

If roots of the equation $x^4-8x^3+bx^2+cx+16=0$ are positive, then
  1. $b=c=8$
  2. $b=-24,c=-32$
  3. $b=24,c=-32$
  4. $b=24,c=32$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a quartic equation with positive roots, by Vieta's formulas, the product of roots is 16 and the sum is 8. If all roots are equal (x=2), the equation is (x-2)^4 = x^4 - 8x^3 + 24x^2 - 32x + 16 = 0. Comparing coefficients, b=24 and c=-32.