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

The integral values of $m$ for which the roots of the equation $mx^2+(2m-1)x+(m-2)=0$ are rational are given by the expression [where $m$ is integer]
  1. $n^2$
  2. $n(n+2)$
  3. $n(n+1)$

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

For the roots of a quadratic equation to be rational, its discriminant (D = b^2 - 4ac) must be a perfect square. Here, D = (2m-1)^2 - 4m(m-2) = 4m^2 - 4m + 1 - 4m^2 + 8m = 4m + 1. For 4m + 1 to be a perfect square, say k^2, then k must be odd, so k = 2n + 1. Then 4m + 1 = (2n + 1)^2 = 4n^2 + 4n + 1, which simplifies to m = n^2 + n = n(n+1).

Multiple choice

If the roots of the equation $ax^2-bx+c=0$ are $\alpha, \beta$, then the roots of the equation $b^2cx^2-ab^2x+a^3=0$ are
  1. $\displaystyle \frac {1}{\alpha^3+\alpha \beta},\displaystyle \frac {1}{\beta^3+\alpha \beta}$
  2. $\displaystyle \frac {1}{\alpha^2+\alpha \beta}, \displaystyle \frac {1}{\beta^2+\alpha \beta}$
  3. $\displaystyle \frac {1}{\alpha^4+\alpha \beta}, \displaystyle \frac {1}{\beta^4+\alpha \beta}$

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

Given the roots of ax^2 - bx + c = 0 are alpha and beta, the transformation of the equation leads to the roots 1/(alpha^2 + alpha*beta) and 1/(beta^2 + alpha*beta).

Multiple choice

If $r$ be the ratio of the roots of the equation $a{ x }^{ 2 }+bx+c=0$, then $\displaystyle \frac { { \left( r+1 \right) }^{ 2 } }{ r } =$
  1. $\displaystyle \frac { { a }^{ 2 } }{ bc } $
  2. $\displaystyle \frac { { b }^{ 2 } }{ ca } $
  3. $\displaystyle \frac { { c }^{ 2 } }{ ab } $

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

Roots are a, ar. Sum = a(1+r) = -b/a. Product = a^2*r = c/a. (r+1)^2 / r = (r^2+2r+1)/r = r + 2 + 1/r = (r^2+1)/r + 2. Using sum/product: a(1+r) = -b/a => 1+r = -b/a^2. This approach is complex; standard identity is (r+1)^2/r = (sum of roots)^2 / (product of roots) = (-b/a)^2 / (c/a) = b^2 / (ac).

Multiple choice

If the equation $ax^2+bx+c=x$ has no real roots, then the equation $a(ax^2+bx+c)^2+b(ax^2+bx+c)+c=x$ will have

  1. four real roots

  2. no real roots

  3. at least two real roots

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

Let f(x) = ax^2 + bx + c. The equation f(x) = x has no real roots, so f(x) - x is never zero. The second equation is f(f(x)) = x. If f(x) = x has no real roots, then f(f(x)) = x implies f(x) must equal x for some x, which is impossible as f(x) - x is never zero.

Multiple choice

If $a$ and $b$ are rational and $b$ is not a perfect square, then the quadratic equation with rational coefficients whose one root is $\displaystyle \frac{1}{a+\sqrt{b}}$ is
  1. $\displaystyle x^{2}-2ax+(a^{2}-b)=0$
  2. $\displaystyle (a^{2}-b)x^{2}-2ax+1=0$
  3. $\displaystyle (a^{2}-b)x^{2}-2bx+1=0$

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

Root x = 1 / (a + sqrt(b)). Then 1/x = a + sqrt(b), so 1/x - a = sqrt(b). Squaring both sides: (1/x - a)^2 = b => 1/x^2 - 2a/x + a^2 = b. Multiply by x^2: 1 - 2ax + a^2x^2 = bx^2. Rearranging: (a^2 - b)x^2 - 2ax + 1 = 0.

Multiple choice

if $ { \alpha }{ 1 },{ \alpha }{ 2 }$ are the roots of the equation $ { x^{ 2 } }+px+q=0 $ and $ { \beta }{ 1 },{ \beta }{ 2 }$ are those of the equation $ { x^{ 2 } }+rx+s=0$ and the system of the equation $ { \alpha }{ 1 }y+{ \alpha }{ 2 }z=0 $ and $ { \beta }{ 1 }y+{ \beta }{ 2 }z=0 $ has non trivial solution, then
  1. $ \dfrac { q^{ 2 } }{ r^{ 2 } } =\dfrac { p}{ s } $
  2. $ \dfrac { p^{ 2 } }{ q^{ 2 } } =\dfrac { r }{ s } $
  3. $ \dfrac { p^{ 2 } }{ s^{ 2 } } =\dfrac { q }{ r } $
  4. $ \dfrac { p^{ 2 } }{ r^{ 2 } } =\dfrac { q }{ s } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A system of linear equations has a non-trivial solution if the determinant of the coefficient matrix is zero: alpha1*beta2 - alpha2*beta1 = 0. This implies alpha1/alpha2 = beta1/beta2. Using the properties of roots for the given quadratic equations, this leads to the relation p^2/r^2 = q/s.