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 the equation for $x$. $13-2(2x+1)=1$
  1. $\displaystyle -\frac{7}{2}$
  2. $\displaystyle -\frac{5}{2}$
  3. $\displaystyle \frac{9}{4}$
  4. $\displaystyle\frac{5}{2}$
  5. $\displaystyle\frac{7}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Expanding the equation: 13 - 4x - 2 = 1. Simplifying gives 11 - 4x = 1, so 4x = 10, which means x = 10/4 = 5/2.

Multiple choice

A quadratic equation whose one root is 3 is
  1. $x^2-6x-5=0$
  2. $x^2+6x-5=0$
  3. $x^2-5x-6=0$
  4. $x^2-5x+6=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If 3 is a root, then substituting x = 3 into the equation must result in 0. For option D, 3^2 - 5(3) + 6 = 9 - 15 + 6 = 0.

Multiple choice

Let $\displaystyle \alpha ,\beta $ be the roots of $\displaystyle { x }^{ 2 }-2x\cos { \phi } +1=0$, then the equation whose roots are $\displaystyle { \alpha }^{ n },{ \beta }^{ n }$ is
  1. $\displaystyle { x }^{ 2 }-2x\cos { n\phi } -1=0$
  2. $\displaystyle { x }^{ 2 }-2x\cos { n\phi } +1=0$
  3. $\displaystyle { x }^{ 2 }-2x\sin { n\phi } +1=0$
  4. $\displaystyle { x }^{ 2 }+2x\sin { n\phi } -1=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Roots are cos(phi) +/- i*sin(phi) = e^(i*phi) and e^(-i*phi). The roots of the new equation are e^(in*phi) and e^(-in*phi). The equation is x^2 - (sum)x + (product) = 0. Sum = 2*cos(n*phi), Product = 1. Equation: x^2 - 2*x*cos(n*phi) + 1 = 0.

Multiple choice

If the ratio of the roots of the equation $ax^{2}+bx+c=0$ is $m : n$, then
  1. $ \displaystyle \frac{m}{n}+\frac{n}{m}=\frac{\ b ^2 }{ac}$
  2. $\sqrt{\left ( \displaystyle \frac{m}{n} \right )}+\sqrt{\left ( \displaystyle \frac{n}{m} \right )}=\displaystyle \frac{\left | b \right |}{\sqrt{c a }}$
  3. $\sqrt{\left ( \displaystyle \frac{m}{n} \right )}+\sqrt{\left (\displaystyle \frac{n}{m} \right )}=\displaystyle \frac{\ b ^2 }{\ c a}$
  4. $\sqrt{\left ( \displaystyle \frac{m}{n} \right )}+\sqrt{\left ( \displaystyle \frac{n}{m} \right )}=\displaystyle \frac{\ a ^2 }{\ b ^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let roots be mr and nr. Sum = (m+n)r = -b/a, Product = mnr^2 = c/a. Squaring the sum and dividing by the product eliminates r, leading to the identity involving square roots of m/n and n/m.

Multiple choice

Which of the following equations has no real roots?
  1. $ x^2 + 4x + 3 \sqrt{2} = 0$
  2. $x^2 + 4x - 3 \sqrt{2} = 0$
  3. $x^2 + 5x + 3 \sqrt{2} = 0$
  4. $3x^2 + 4 \sqrt{3}x + 4 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

An equation has no real roots if the discriminant D < 0. For A: D = 4^2 - 4(1)(3*sqrt(2)) = 16 - 12*sqrt(2) = 16 - 16.97 < 0.

Multiple choice

If $\displaystyle \alpha ,\beta $ are the roots of a quadratic equation such that $\displaystyle \alpha +\beta =24$ and $\displaystyle \alpha -\beta =8$, the quadratic equation is
  1. $\displaystyle { x }^{ 2 }-24x+128=0$
  2. $\displaystyle { x }^{ 2 }-8x+16=0$
  3. $\displaystyle { x }^{ 2 }-24x+12=0$
  4. $None\ of\ the\ above$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given alpha + beta = 24 and alpha - beta = 8, we find 2*alpha = 32 -> alpha = 16, and beta = 8. The equation is x^2 - (sum)x + (product) = 0, which is x^2 - 24x + 128 = 0.