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

Solve following equation : $x^2 + 3 = 0$
  1. $\pm 3$
  2. $\pm \sqrt{3}$
  3. $\pm i\sqrt{3}$
  4. $\pm {3}i$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solving x^2 + 3 = 0 leads to x^2 = -3. Taking the square root of both sides gives x = +/- sqrt(-3), which is +/- i*sqrt(3).

Multiple choice

Solve the following quadratic equation by completing square. $2y^{2}\, +\, 5y\, +\, 1\, =\, 0$, then $y\, =\, \displaystyle \frac{-5\, \pm\, \sqrt{23}}{2}$. State true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 2y^2 + 5y + 1 = 0, the quadratic formula is y = (-b +/- sqrt(b^2 - 4ac)) / 2a. Here a=2, b=5, c=1. y = (-5 +/- sqrt(25 - 8)) / 4 = (-5 +/- sqrt(17)) / 4. The statement provides sqrt(23) and denominator 2, which is incorrect.

Multiple choice

Find the roots of each of the following quadratic equations by the method of completing the squares $(5z + 2a) (3z + 4b) = 8ab$
  1. Roots are $\displaystyle 1 , - \frac{a + 10b}{11}$
  2. Roots are $\displaystyle 7 , - \frac{a - 15b}{17}$
  3. Roots are $\displaystyle 0 , - \frac{6a + 20b}{15}$
  4. Roots are $\displaystyle -1 , - \frac{7a - 10b}{17}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Expanding the equation (5z + 2a)(3z + 4b) = 8ab gives 15z^2 + 20bz + 6az + 8ab = 8ab, which simplifies to 15z^2 + (20b + 6a)z = 0. Factoring out z yields z(15z + 6a + 20b) = 0, so the roots are z = 0 and z = -(6a + 20b)/15.

Multiple choice

Consider the quadratic equations $\displaystyle ax^{2}+2bx+c=0$ and $\displaystyle \left ( a+c \right )\left ( ax^{2}+2bx+c \right )-2\left ( ac-b^{2} \right )\left ( x^{2}+1 \right )=0$ If the roots of one are real (complex) then the roots of the other are complex (real.)

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The second equation is a transformation of the first. The relationship between the discriminants shows that if one has real roots, the other has complex roots, and vice versa.

Multiple choice

Find the roots of the following quadratic equations, if they exist, by the method of completing the square. $\displaystyle 2x^2 - 7x + 3 = 0$
  1. The roots of the given quadratic equation are $\displaystyle \frac{1}{3}$ and $1$.
  2. The roots of the given quadratic equation are $\displaystyle \frac{2}{5}$ and $5$.
  3. The roots of the given quadratic equation are $\displaystyle \frac{1}{2}$ and $3$.
  4. The roots of the given quadratic equation are $\displaystyle \frac{2}{7}$ and $4$.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

2x^2 - 7x + 3 = 0. Divide by 2: x^2 - 3.5x + 1.5 = 0. (x - 1.75)^2 = 1.75^2 - 1.5 = 3.0625 - 1.5 = 1.5625. x - 1.75 = +/- 1.25. x = 3 or x = 0.5.

Multiple choice

Find the roots of each of the following quadratic equations by the method of completing the squares $\sqrt 5 x^2 + 9x + 4 \sqrt 5 = 0$
  1. $\displaystyle - \sqrt 7, \frac{-17}{\sqrt 3 }$
  2. $\displaystyle - \sqrt 5, \frac{-4}{\sqrt 5 }$
  3. $\displaystyle - \sqrt 5, \frac{-14}{\sqrt 3 }$
  4. $\displaystyle - \sqrt 7, \frac{-13}{\sqrt 5 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To solve sqrt(5)x^2 + 9x + 4sqrt(5) = 0, we can factor by grouping. We look for two numbers that multiply to (sqrt(5) * 4sqrt(5)) = 20 and add to 9. These are 5 and 4. The equation becomes sqrt(5)x^2 + 5x + 4x + 4sqrt(5) = 0. Factoring gives sqrt(5)x(x + sqrt(5)) + 4(x + sqrt(5)) = 0, which leads to (sqrt(5)x + 4)(x + sqrt(5)) = 0. The roots are -sqrt(5) and -4/sqrt(5).

Multiple choice

Solve the following equations: $xy + x + y = 23$, $xz + x + z = 41$, $yz + y + z = 27$.
  1. $x=4, -2; y=2; 6; z=6, -5$
  2. $x=2, -4; y=2,4 ;z=2, -6$
  3. $x = 5, -7; y = 3, -5; z = 6, -8$
  4. $x=3, 4; y=2, -5; z=2,-7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given xy+x+y=23, xz+x+z=41, yz+y+z=27. Adding 1 to each side: (x+1)(y+1)=24, (x+1)(z+1)=42, (y+1)(z+1)=28. Let X=x+1, Y=y+1, Z=z+1. Then XY=24, XZ=42, YZ=28. Multiplying gives (XYZ)^2 = 24*42*28 = 28224, so XYZ = 168. Then Z = 168/24 = 7, Y = 168/42 = 4, X = 168/28 = 6. Thus x=5, y=3, z=6. The negative roots follow from the quadratic nature.

Multiple choice

Find all values of $a$ for which both roots of the equation ${ x }^{ 2 }-6ax+2-2a+9{ a }^{ 2 }=0$ are greater than $3$.
  1. For all $a\in \left( { 11 }/{ 9 },+\infty \right) $
  2. For all $a\in \left( { 22 }/{ 9 },+\infty \right) $
  3. For all $a\in \left( { 11 }/{ 3 },+\infty \right) $
  4. For all $a\in \left( { 1 }/{ 9 },+\infty \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For both roots > 3: 1) Discriminant >= 0: 36a^2 - 4(9a^2-2a+2) >= 0 -> 8a - 8 >= 0 -> a >= 1. 2) Vertex > 3: -b/2a = 6a/2 = 3a > 3 -> a > 1. 3) f(3) > 0: 9 - 18a + 9a^2 - 2a + 2 > 0 -> 9a^2 - 20a + 11 > 0 -> (9a-11)(a-1) > 0. For a > 1, this holds if a > 11/9.