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 the sides $a,\ b,\ c$ of a triangle $ABC$ are the roots of the equation $x^{3}-13x^{2}+54x-72=0$, then the value of $\displaystyle \frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}$ is equal to?
  1. $\displaystyle \frac{169}{144}$
  2. $\displaystyle \frac{61}{72}$
  3. $\displaystyle \frac{61}{144}$
  4. $\displaystyle \frac{169}{72}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the roots be a, b, c. From Vieta's formulas, a+b+c = 13, ab+bc+ca = 54, abc = 72. Using the cosine rule, cosA = (b^2+c^2-a^2)/(2bc). The expression (cosA/a + cosB/b + cosC/c) simplifies to (a^2+b^2+c^2)/(2abc). Using (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca), we get 169 = a^2+b^2+c^2 + 108, so a^2+b^2+c^2 = 61. The result is 61 / (2 * 72) = 61/144.

Multiple choice

lf $\alpha,\ \beta$ are the roots of $4{x}^{2}+7{x}+2=0$, then the equation whose roots are $\alpha^{2},\ \beta^{2}$ is
  1. $16{x}^{2}-33{x}+4=0$
  2. $16{x}^{2}+33{x}+4=0$
  3. $4{x}^{2}-49{x}+2=0$
  4. $16{x}^{2}-4{x}+2=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If alpha and beta are roots of 4x^2 + 7x + 2 = 0, then alpha + beta = -7/4 and alpha*beta = 2/4 = 1/2. The new equation with roots alpha^2 and beta^2 is x^2 - (alpha^2 + beta^2)x + (alpha*beta)^2 = 0. Since alpha^2 + beta^2 = (alpha + beta)^2 - 2(alpha*beta) = 49/16 - 1 = 33/16, the equation is x^2 - (33/16)x + 1/4 = 0, which simplifies to 16x^2 - 33x + 4 = 0.

Multiple choice

If $ \alpha, \beta $ are roots of the equation $ 2x^{2}+6x+b=0&gt;(b<0)$, then $ \dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha} $ is less than
  1. $2$
  2. $-2$
  3. $18$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 2x^2 + 6x + b = 0, the sum of roots alpha + beta = -6/2 = -3 and product alpha*beta = b/2. The expression alpha/beta + beta/alpha = (alpha^2 + beta^2) / (alpha*beta) = ((alpha + beta)^2 - 2*alpha*beta) / (alpha*beta) = (9 - b) / (b/2) = 18/b - 2. Since b < 0, 18/b is negative, so 18/b - 2 < -2.

Multiple choice

If $\displaystyle x^{2}-2r\times p_{r}x+r=0;:r=1,2,3$ are the quadratc equations of which each pair has exaclty one root common then the number of solutions of the triplet $(p_{1},p_{2},p_{3})$ is
  1. $2$
  2. $1$
  3. $9$
  4. $27$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For two quadratics to have exactly one common root, the condition is (c1a2 - c2a1)^2 = (a1b2 - a2b1)(b1c2 - b2c1). Applying this to the pairs of equations provided leads to a system that restricts the possible values of p1, p2, p3.

Multiple choice

Solve the equation: $x^2 - 5x + 6 \geq 0 $.
  1. $ x \in (- \infty , 2] \cup [3, \infty)$
  2. $ x \in [2,3] $
  3. $ x \in (- \infty , -1] \cup [6, \infty)$
  4. $ x \in [-1,6] $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

x^2 - 5x + 6 >= 0 factors to (x-2)(x-3) >= 0. The inequality holds when x <= 2 or x >= 3.