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 roots of the quadratic equation $x^2-2x-3=0$ are $-1$ and $3$. The graph of $x^2-2x-3$ will pass through which of the following points.
  1. $(2,0) and (3,0)$
  2. $(-1,0) and (3,0)$
  3. $(0,0) and (1,0)$

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

The roots of a quadratic equation are the x-values where the graph intersects the x-axis, meaning the y-coordinate is 0. Since the roots are -1 and 3, the graph must pass through the points (-1, 0) and (3, 0).

Multiple choice

The roots of the quadratic equation $3x^2-2x-1=0$ are 1 and $-\dfrac{1}{3}$. From this information, determine which of the following options will be the zeros of the quadratic polynomial $3x^2-2x-1$
  1. $x= 0$ and $x=1$
  2. $x=3$ and $x=2$
  3. $x=1$ and $x = -\dfrac{1}{3}$

  4. Cannot be determined.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The zeros of a polynomial are the values of x for which the polynomial equals zero. Since the roots of the equation 3x^2 - 2x - 1 = 0 are 1 and -1/3, these are by definition the zeros of the polynomial 3x^2 - 2x - 1.

Multiple choice

If the roots of the equation $x^{2} + a^{2} = 8x + 6a$ are real, then:
  1. $a \epsilon \left [ 2,8 \right ]$
  2. $ a \epsilon \left [ -2,8 \right ]$
  3. $a \epsilon (2,8)$
  4. $ a \epsilon (-2,8)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For real roots, the discriminant D >= 0. Equation: x^2 - 8x + (a^2 - 6a) = 0. D = (-8)^2 - 4(1)(a^2 - 6a) = 64 - 4a^2 + 24a. 64 + 24a - 4a^2 >= 0 implies a^2 - 6a - 16 <= 0. Factoring: (a - 8)(a + 2) <= 0. This holds for a in [-2, 8].

Multiple choice

If the roots of the equation $x^2 + 2 (3a + 5)x + 2(9a^2 + 25) = 0$ are real, then what will be the value of 'a'?
  1. a > $ \frac { 5 }{ 3 }$
  2. a < $\frac { 5 }{ 3 }$
  3. a = $\frac { 5 }{ 3 }$

  4. a > 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For real roots, the discriminant D = b^2 - 4ac >= 0. Here, D = [2(3a+5)]^2 - 4(1)(2(9a^2+25)) = 4(9a^2 + 30a + 25) - 8(9a^2 + 25) = 36a^2 + 120a + 100 - 72a^2 - 200 = -36a^2 + 120a - 100. Setting -4(9a^2 - 30a + 25) = -4(3a-5)^2 >= 0. Since a square is always non-negative, this is only >= 0 when 3a-5 = 0, so a = 5/3.

Multiple choice

If $\alpha $ , $\beta $ are the roots of quadratic equation ${x^2} + px + q = 0$ and $\gamma $, $\delta $ are the roots of ${x^2} + px - r = 0$ then $\left( {\alpha - \gamma } \right)$.$\left( {\alpha - \delta } \right)$ is equal to:
  1. $q + r$
  2. $q - r$
  3. $-(q + r)$
  4. $-(p + q + r)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given the roots, the expressions are (alpha - gamma)(alpha - delta) = alpha^2 - alpha(gamma + delta) + gamma*delta. Using Vieta's formulas, gamma + delta = -p and gamma*delta = -r. Substituting these into the quadratic x^2 + px + q = 0, we get alpha^2 + p*alpha + q = 0, so alpha^2 = -p*alpha - q. Substituting gives (-p*alpha - q) - alpha(-p) - r = -q - r = -(q + r).