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 real roots of the equation $(\sqrt{3}+1)^{2x}+(\sqrt{3}-1)^{2x}=2^{3x}$ is equal to
  1. $0$
  2. $1$
  3. $2$
  4. more than $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Divide by (sqrt(3)+1)^(2x). The equation becomes 1 + ((sqrt(3)-1)/(sqrt(3)+1))^(2x) = (2/(sqrt(3)+1))^(2x). This simplifies to 1 + (2-sqrt(3))^x = (sqrt(3)-1)^(2x). Testing x=1 gives 1 + 2-sqrt(3) = 3-2sqrt(3)+1, which is not equal. Testing x=0 gives 1+1=1 (false). Graphical analysis shows only one intersection.

Multiple choice

The roots of the equation $ \displaystyle \sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6} $ are
  1. $ \displaystyle \frac{3}{13},\frac{2}{13} $
  2. $ \displaystyle \frac{13}{6},\frac{11}{6} $
  3. $ \displaystyle \frac{9}{13},\frac{4}{13} $

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

Let y = sqrt(x/(1-x)). Then y + 1/y = 13/6. Solving 6y^2 - 13y + 6 = 0 gives y = 2/3 or 3/2. If sqrt(x/(1-x)) = 2/3, then x/(1-x) = 4/9, so 9x = 4 - 4x, 13x = 4, x = 4/13. If y = 3/2, x = 9/13.

Multiple choice

Roots of the equation $x^{2}-6x-\sqrt{x^{2}-6x-3}=5$ are
  1. $1, 7$
  2. $-1, -7$
  3. $1$, $-1$
  4. $-1$, $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let y = sqrt(x^2 - 6x - 3). Then y^2 = x^2 - 6x - 3, so x^2 - 6x = y^2 + 3. The equation becomes y^2 + 3 - y = 5, or y^2 - y - 2 = 0. Solving gives y = 2 or y = -1 (discarded). x^2 - 6x - 3 = 4 leads to x^2 - 6x - 7 = 0, so (x-7)(x+1) = 0.

Multiple choice

Number of rational roots of the equation $\sqrt{5x^{2}-6x+8}-\sqrt{5x^{2}-6x-7}=1$ is/are
  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let y = 5x^2 - 6x. The equation becomes sqrt(y+8) - sqrt(y-7) = 1. Squaring both sides gives (y+8) + (y-7) - 2*sqrt((y+8)(y-7)) = 1, which simplifies to 2y + 1 - 1 = 2*sqrt(y^2 + y - 56), or y = sqrt(y^2 + y - 56). Squaring again, y^2 = y^2 + y - 56, so y = 56. Substituting back, 5x^2 - 6x - 56 = 0. Solving via quadratic formula, x = (6 +/- sqrt(36 - 4*5*(-56)))/10 = (6 +/- sqrt(1156))/10 = (6 +/- 34)/10. The roots are 4 and -2.8. Both are rational.

Multiple choice

Solve the following equations: $\dfrac {\sqrt {x} - \sqrt {y}}{\sqrt {x} + \sqrt {y}} + \dfrac {\sqrt {x} + \sqrt {y}}{\sqrt {x} - \sqrt {y}} = \dfrac {17}{4}$, $x^{2} + y^{2} = 706$
  1. $x=\pm 9, y=\pm 25$
  2. $x=\pm 25, y=\pm 9$
  3. $x=\pm 36, y=\pm 49$
  4. $x=\pm 9, y=\pm 36$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let u = sqrt(x), v = sqrt(y). The equation is (u-v)/(u+v) + (u+v)/(u-v) = 17/4. This simplifies to 2(u^2+v^2)/(u^2-v^2) = 17/4. 8(x+y) = 17(x-y). Also x^2+y^2=706. Testing options: For x=25, y=9: 25^2+9^2 = 625+81 = 706. Check first eq: (5-3)/(5+3) + (5+3)/(5-3) = 2/8 + 8/2 = 0.25 + 4 = 4.25 = 17/4. Correct.