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

Let $p,q\ne 0$ be constant and let $\alpha,\beta$ be the roots of the equation $x^2+px+q=0$. Then find the roots of the equation $qx^2+px+1=0$ are
  1. $\alpha,\beta$
  2. $\dfrac{1}{\alpha},\dfrac{1}{\beta}$
  3. $\alpha^2,\beta^2$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If alpha and beta are roots of x^2+px+q=0, then alpha+beta = -p and alpha*beta = q. For qx^2+px+1=0, the roots are ( -p +/- sqrt(p^2 - 4q) ) / 2q. This simplifies to 1/alpha and 1/beta.

Multiple choice

Product of the roots of the equation $x^{2}+3|x|-28=0$ is
  1. $0$
  2. $-16$
  3. $16$
  4. $-28$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By substituting y = |x| (where y must be non-negative), the equation becomes y^2 + 3y - 28 = 0, which factors into (y + 7)(y - 4) = 0. Since y cannot be negative, we have y = 4, which gives the real roots x = 4 and x = -4. The product of these roots is 4 * (-4) = -16.

Multiple choice

If roots of the equation $ { ax }^{ 2 }+bx+c=0$ are $ \dfrac { \alpha }{ \alpha -1 } $ and $ \dfrac { \alpha+1 }{ \alpha } $, then $ { \left( a+b+c \right) }^{ 2 }={ b }^{ 2 }+4ac$

  1. True

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

The roots are r1 = a/(a-1) and r2 = (a+1)/a. Sum = r1+r2 = -b/a, Product = r1*r2 = c/a. The identity (a+b+c)^2 = b^2+4ac is not generally true for these roots. Testing with a simple case: if a=1, b=-2, c=0, roots are 2 and 2, but the equation doesn't hold.

Multiple choice

If the equation $a{x^2} + bx + 6 = 0$ does not have two distinct real roots, then the least value of $3a + b$ is
  1. $3$
  2. $ - 3$
  3. $2$
  4. $ - 2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the quadratic ax^2 + bx + 6 = 0 to not have two distinct real roots, the discriminant D = b^2 - 4(a)(6) must be <= 0. So b^2 <= 24a. We want the minimum value of 3a + b. Testing values, if a=1, b^2 <= 24, so b can be -4. 3(1) - 4 = -1. If a=1/3, b^2 <= 8, b approx -2.8. If a=1/6, b^2 <= 4, b = -2. 3(1/6) - 2 = 0.5 - 2 = -1.5. Checking the option -2: if 3a+b = -2, b = -2-3a. (-2-3a)^2 <= 24a -> 4 + 12a + 9a^2 <= 24a -> 9a^2 - 12a + 4 <= 0 -> (3a-2)^2 <= 0. This holds for a=2/3, b=-4.

Multiple choice

The quadratic equation ${x}^{2}-9x+3=0$ has roots $\alpha$ and $\beta$. If ${x}^{2}-bx-c=0$ has roots ${\alpha}^{2}$ and ${\beta}^{2}$, then $(b,c)$ is
  1. $\left(75,-9\right)$
  2. $\left(-75,9\right)$
  3. $\left(-87,4\right)$
  4. $\left(-87,9\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the given equation, the sum of roots is alpha + beta = 9 and the product is alpha*beta = 3. The new roots are alpha^2 and beta^2, so their sum is alpha^2 + beta^2 = (alpha + beta)^2 - 2*alpha*beta = 81 - 6 = 75, and their product is alpha^2 * beta^2 = 9. Comparing the new equation x^2 - 75x + 9 = 0 with x^2 - bx - c = 0 gives b = 75 and c = -9.

Multiple choice

If p and q are the roots of the equation $x ^ { 2 } + p x + q = 0 ,$ then
  1. $\mathrm { p } = 1 , \mathrm { q } = - 2$
  2. $p = 0 , q = 1$
  3. $\mathrm { p } = - 2 , \mathrm { q } = 0$
  4. $p = - 2 , q = 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since p and q are the roots of the quadratic equation x^2 + px + q = 0, the sum of the roots is p + q = -p, which simplifies to q = -2p. The product of the roots is p * q = q. If q is not zero, dividing both sides by q gives p = 1, which then yields q = -2.

Multiple choice

The roots of the equation $x^4-1=0$, are?
  1. $1, 1, i, -i$
  2. $1, -1, i, -i$
  3. $1, -1, \omega, \omega^2$

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

x^4 - 1 = 0 => (x^2 - 1)(x^2 + 1) = 0. (x-1)(x+1)(x-i)(x+i) = 0. Roots are 1, -1, i, -i.