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

Form a quadratic equation whose roots are: $3, 4$
  1. $x^2-7x+12=0$
  2. $x^2+7x+12=0$
  3. $x^2-7x-12=0$

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

A quadratic equation with roots r1, r2 is x^2 - (r1+r2)x + (r1*r2) = 0. Here r1=3, r2=4. Sum = 7, product = 12. Equation: x^2 - 7x + 12 = 0.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $x^2-3x+2=0$, form a equation whose roots are of $-\alpha $ and $-\beta$.
  1. $x^2+3x+2=0$
  2. $x^2-3x+2=0$
  3. $x^2+3x-2=0$

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

If roots are alpha and beta, the equation is x^2 - (alpha+beta)x + alpha*beta = 0. Here alpha+beta = 3 and alpha*beta = 2. New roots are -alpha and -beta. New sum = -(alpha+beta) = -3. New product = (-alpha)*(-beta) = alpha*beta = 2. New equation is x^2 - (-3)x + 2 = 0, which is x^2 + 3x + 2 = 0.

Multiple choice

If $\alpha$, $\beta$ are the roots of the equation $a{ x }^{ 2 }+bx+b=0$, then what is the value of $\sqrt { \dfrac { \alpha }{ \beta } } +\sqrt { \dfrac { \beta }{ \alpha } } +\sqrt { \dfrac { b }{ a } } $?
  1. $-1$
  2. $0$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For ax^2 + bx + b = 0, alpha + beta = -b/a and alpha*beta = b/a. The expression is sqrt(alpha/beta) + sqrt(beta/alpha) + sqrt(b/a) = (alpha+beta)/sqrt(alpha*beta) + sqrt(b/a) = (-b/a) / sqrt(b/a) + sqrt(b/a) = -sqrt(b/a) + sqrt(b/a) = 0.

Multiple choice

Let $\alpha ,\beta $ be the roots of ${ x }^{ 2 }+3x+5=0$, then the equation whose roots are $-\cfrac { 1 }{ \alpha } $ and $-\cfrac { 1 }{ \beta } $ is
  1. $5{ x }^{ 2 }+3x-4=0$
  2. $5{ x }^{ 2 }-3x+4=0$
  3. $5{ x }^{ 2 }+3x-1=0$
  4. $5{ x }^{ 2 }-3x+1=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If roots are -1/alpha and -1/beta, we substitute x = -1/y into the original equation x^2 + 3x + 5 = 0. This gives (-1/y)^2 + 3(-1/y) + 5 = 0, which is 1/y^2 - 3/y + 5 = 0. Multiplying by y^2 results in 5y^2 - 3y + 1 = 0.

Multiple choice

$\alpha,\beta$ be the roots of the equation ${ x }^{ 2 }-3x+a=0$ and $\gamma ,\delta $ the roots of ${ x }^{ 2 }-12x+b=0$ and numbers $\alpha ,\beta ,\gamma ,\delta $ (in this order) form an increasing G.P, then
  1. $a=3,b=12$
  2. $a=12,b=3$
  3. $a=2,b=32$
  4. $a=4,b=16$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the four increasing GP terms be t, tr, tr^2, and tr^3. Since the first pair sums to 3 and the second pair sums to 12, r^2 = 4, so r = 2 and the terms are 1, 2, 4, 8, giving a = 2 and b = 32.

Multiple choice

Choose the correct answer from the alternatives given. If a and $\beta$ be the roots of the equation $ax^2 \, + \, bx \, + \, c \, = \, 0,$ then find the values of $\dfrac{\alpha^2}{\beta} \, + \, \dfrac{\beta^2}{\alpha}.$
  1. $\displaystyle \frac{ab \, - \, b^2 \, c}{2b^2 \, c}.$
  2. $\displaystyle \frac{3bc \, - \, a^2}{2b^2 \, c}.$
  3. $\displaystyle \frac{3ac \, - \, b^2}{a^3 \, c}.$
  4. $\displaystyle \frac{3abc \, - \, b^3}{a^2 \, c}.$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

alpha^2/beta + beta^2/alpha = (alpha^3 + beta^3)/(alpha*beta). alpha+beta = -b/a, alpha*beta = c/a. alpha^3+beta^3 = (alpha+beta)^3 - 3*alpha*beta*(alpha+beta) = (-b/a)^3 - 3*(c/a)*(-b/a) = -b^3/a^3 + 3bc/a^2 = (3abc - b^3)/a^3. Dividing by c/a gives (3abc - b^3)/(a^2*c).

Multiple choice

If $\alpha \, and \, \beta$ are the roots of the equation $6x^2 \, - \, 6x \, + \, 1 \, = \,0$ then $1/2 \left ( p \, + \, q\alpha \, + \, r\alpha ^2 \, + \, s\alpha ^3 \right ) \, + \, 1/2\left ( p \, + \, q\beta \, + \, r\beta ^2 \, + \,s\beta ^3 \right )$ is $\dfrac{p}{1} \, + \,\dfrac{q}{2} \, + \, \dfrac{r}{3} \, + \, \dfrac{s}{4}.$

  1. True

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

This is a known identity related to the power sums of roots of a quadratic equation. The expression evaluates correctly based on the Newton's sums for the given polynomial.