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

If the ratio of the roots of equation $\displaystyle x^{2}+px+q=0$ be equal to the ratio of the roots of $\displaystyle x^{2}+lx+m=0$ then
  1. $\displaystyle p^{2}m=q^{2}l$
  2. $\displaystyle pm^{2}=q^{2}l$
  3. $\displaystyle p^{2}l=q^{2}m$
  4. $\displaystyle p^{2}m=l^{2}q$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let roots of first be ar, br and second be cr, dr. Ratio r = a/b = c/d. Sums are -p, -l; products are q, m. (a+b)^2 / ab = (c+d)^2 / cd => p^2 / q = l^2 / m => p^2 m = l^2 q.

Multiple choice

If p and q are non-zero constants the equation $\displaystyle x^{2}+px+q=0$ has roots$\displaystyle \alpha $ and $\displaystyle \beta $ the equation $\displaystyle qx^{2}+px+1=0$ has roots
  1. $\displaystyle \alpha $ and $\displaystyle \frac{1}{\beta }$
  2. $\displaystyle \frac{1}{\alpha }$ and $\displaystyle \beta $
  3. $\displaystyle \frac{1}{\alpha }$ and $\displaystyle \frac{1}{\beta }$

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

If x^2 + px + q = 0 has roots alpha and beta, then alpha * beta = q and alpha + beta = -p. For qx^2 + px + 1 = 0, the product of roots is 1/q = 1/(alpha * beta) and sum of roots is -p/q = (alpha + beta) / (alpha * beta) = 1/alpha + 1/beta. The roots are 1/alpha and 1/beta.

Multiple choice

The roots of the equation $x^2-8x -16= 0$

  1. are both positive

  2. are both negative
  3. one root lies in the interval $(-1, 0)$
  4. lie in the interval $(-2, 10)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For x^2 - 8x - 16 = 0, the roots are x = [8 +/- sqrt(64 - 4(1)(-16))] / 2 = [8 +/- sqrt(128)] / 2 = 4 +/- 4*sqrt(2). Roots are approx 4 + 5.65 = 9.65 and 4 - 5.65 = -1.65. Both roots lie in the interval (-2, 10).

Multiple choice

In a triangle $ABC, 3\cos A+2=0$. Then the quadratic equation with roots are $\sin A, \cos A$ is
  1. $9x^2 - 3x\left( \sqrt { 5 } -2 \right) -2\sqrt { 5 } =0$
  2. $9x^2 - 3x\left( \sqrt { 5 } - 2 \right) +2\sqrt { 5 } =0$
  3. $9x^2 - 3x\left( \sqrt { 7 } - 2 \right) +2\sqrt { 7 } =0$
  4. $9x^2 - 3x\left( \sqrt { 11 } - 2 \right) +2\sqrt { 11 }=0 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

3 cos A = -2, cos A = -2/3. sin A = sqrt(1 - 4/9) = sqrt(5)/3. Sum of roots = -2/3 + sqrt(5)/3 = (sqrt(5)-2)/3. Product = -2*sqrt(5)/9. Equation: x^2 - (sum)x + product = 0. x^2 - ((sqrt(5)-2)/3)x - 2*sqrt(5)/9 = 0. Multiply by 9: 9x^2 - 3x(sqrt(5)-2) - 2*sqrt(5) = 0.

Multiple choice

Solve the equation by the square root property. ${x}^{2} = 8$
  1. $\left\{ \sqrt { 8 } \right\} $
  2. $\left\{ 64 \right\} $
  3. $\left\{ \pm 2\sqrt { 2 } \right\} $
  4. $\left\{ 4 \right\} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

x^2 = 8 implies x = +sqrt(8) or x = -sqrt(8). sqrt(8) = 2*sqrt(2). Thus, the solutions are {2*sqrt(2), -2*sqrt(2)}.

Multiple choice

Solve the equation using the quadratic formula $2=-\displaystyle\frac { 10 }{ x } -\displaystyle\frac { 5 }{ { x }^{ 2 } } $
  1. $\left\{ \displaystyle\frac { -5\pm \sqrt { 15 } }{ 2 } \right\} $
  2. $\left\{ \displaystyle\frac { -5\pm \sqrt { 15 } }{ 4 } \right\} $
  3. $\left\{ \displaystyle\frac { -5\pm \sqrt { 35 } }{ 2 } \right\} $
  4. $\left\{ \displaystyle\frac { -10\pm \sqrt { 15 } }{ 2 } \right\} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Multiply by x^2: 2x^2 = -10x - 5, which is 2x^2 + 10x + 5 = 0. Using the quadratic formula x = (-b +/- sqrt(b^2 - 4ac)) / 2a: x = (-10 +/- sqrt(100 - 40)) / 4 = (-10 +/- sqrt(60)) / 4 = (-10 +/- 2*sqrt(15)) / 4 = (-5 +/- sqrt(15)) / 2.

Multiple choice

The roots of the equation ${2x^{2 }+ 3x + 2} = 0$ are

  1. Real, rational, and equal

  2. Real,rational and unequal

  3. Real, irrational, and unequal

  4. non real (imaginary)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the equation 2x^2 + 3x + 2 = 0, the discriminant D = b^2 - 4ac = 3^2 - 4(2)(2) = 9 - 16 = -7. Since the discriminant is negative, the roots are non-real (imaginary).