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

Number of roots of a quadratic equation are:
  1. $1$
  2. $2$
  3. $3$

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

A quadratic equation is of the form ax^2 + bx + c = 0, which by the Fundamental Theorem of Algebra has exactly two roots (counting multiplicity).

Multiple choice

The roots of the equation $\displaystyle 2{ y }^{ 2 }+y-2=0$ are
  1. $\displaystyle \frac { -1-\sqrt { 17 } }{ 2 } ,\frac { -1-\sqrt { 17 } }{ 2 } $
  2. $\displaystyle \frac {- 1-\sqrt { 17 } }{ 4 } ,\frac { -1+\sqrt { 17 } }{ 4 } $
  3. $\displaystyle -1-\sqrt { 17 } ,-1+\sqrt { 17 } $

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

For 2y^2 + y - 2 = 0, use the quadratic formula y = (-b +/- sqrt(b^2 - 4ac)) / 2a. Here a=2, b=1, c=-2. y = (-1 +/- sqrt(1 - 4*2*(-2))) / 4 = (-1 +/- sqrt(17)) / 4.

Multiple choice

Let $a, b$ and $c$ be three real numbers, such that $a+2b+4c=0$. Then the equation $a{ x }^{ 2 }+bx+c=0$

  1. has both the roots complex
  2. has its roots lying within $-1< x< 0$
  3. has one of the roots equal to $\cfrac{1}{2}$
  4. has its roots lying within $2< x< 6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given a + 2b + 4c = 0, we can write c = -a/4 - b/2. Substituting this into the quadratic equation ax^2 + bx + c = 0 gives ax^2 + bx - (a/4 + b/2) = 0. Testing x = 1/2: a(1/4) + b(1/2) - a/4 - b/2 = 0, which confirms x = 1/2 is a root.

Multiple choice

If one root of the quadratic equation $ax^2 + bx + c = 0$ is equal to $n^{th}$ power of the other root, then the value of: $a^{\frac{n}{n+1}}c^{\frac{1}{n+1}}+ c^{\frac{n}{n+1}}a^{\frac{1}{n+1}}$ is equal to
  1. $b$
  2. $-b$
  3. $\dfrac{1}{b^{n+1}}$
  4. $-b^{n+1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If roots are r and r^n, then r * r^n = c/a and r + r^n = -b/a. The expression a^(n/(n+1)) * c^(1/(n+1)) + c^(n/(n+1)) * a^(1/(n+1)) simplifies to (a*c)^(1/(n+1)) * (a^(n/(n+1)) + c^(n/(n+1))). This is a known identity related to the roots of ax^2+bx+c=0.