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

If $a,b,c,r \in R$, then the equation $(x^2+ax-3b)(x^2-cx+b)(x^2-dx+2b)=0$ has
  1. $6$ real roots
  2. at least $2$ real roots
  3. $4$ real roots
  4. $3$ real roots
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If b < 0, the second quadratic has discriminant c^2 - 4b > 0, so it has two real roots. If b >= 0, the first quadratic has discriminant a^2 + 12b >= 0, giving real roots under the usual multiplicity convention. Thus, at least two real roots are guaranteed.

Multiple choice

The coefficient of $x$ in the equation $x^2+px+q=0$ was wrongly written as $17$ in place of $13$ and the roots thus found was $-2$ and $-15$. Then the roots of the correct equation are
  1. $-3,10$
  2. $-3,-10$
  3. $3,-10$

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

Roots -2 and -15 imply (x+2)(x+15) = x^2 + 17x + 30 = 0. The wrong coefficient was 17, so the correct equation is x^2 + 13x + 30 = 0. Factoring gives (x+3)(x+10) = 0, so roots are -3 and -10.

Multiple choice

$(a^{4}-1)x^{2}-(a^{2}+1)(sin^{-1}sin^{3} 2)x+(cos^{-1}cos2)(a^{2}-1) =0$. .Find the set of values of a so that above equation have roots of opposite in sign.

  1. [-2 , 3]

  2. [-1 , 1]

  3. [-2 ,1]
  4. $\phi $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For roots to be opposite in sign, the product of roots (c/a) must be negative. The equation is (a^4-1)x^2 - (a^2+1)(sin^-1(sin^3(2)))x + (cos^-1(cos 2))(a^2-1) = 0. Since sin^-1(sin 2) = pi-2 and cos^-1(cos 2) = 2, the constant term is 2(a^2-1). For the product to be negative, 2(a^2-1)/(a^4-1) < 0. Simplifying, 2/(a^2+1) < 0, which is impossible for any real a.

Multiple choice

The number of integral values of $m$ less than $50$ such that the roots of the quadratic equation $mx^{2}+\left ( 2m-1 \right )x+\left ( m-2 \right )=0$ are rational are
  1. $6$
  2. $7$
  3. $13$
  4. $14$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For roots to be rational, the discriminant D = (2m-1)^2 - 4(m)(m-2) must be a perfect square. D = 4m^2 - 4m + 1 - 4m^2 + 8m = 4m + 1. For 4m+1 to be a perfect square, let 4m+1 = k^2. Since 4m+1 is odd, k must be odd. k^2 = 1, 9, 25, 49, 81, 121, 169, 225, 289. 4m+1 = k^2 => m = (k^2-1)/4. For m < 50, k^2 < 201. Possible odd k^2 values: 1, 9, 25, 49, 81, 121, 169. This gives m = 0, 2, 6, 12, 20, 30, 42. There are 7 values, but m cannot be 0 for a quadratic equation. So 6 values.

Multiple choice

If the equation $\displaystyle \frac{x^{2}-bx}{ax-c}=\frac{m-1}{m+1}$ has roots equal in magnitude but opposite in sign, then $m=$
  1. $\displaystyle \frac{a+b}{a-b}$
  2. $\displaystyle \frac{a-b}{a-b}$
  3. $\displaystyle \frac{a-b}{a+b}$

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

For the equation to have roots equal in magnitude but opposite in sign, the sum of the roots must be zero. Given the equation (x^2 - bx) / (ax - c) = (m-1) / (m+1), cross-multiply to get (m+1)(x^2 - bx) = (m-1)(ax - c). Expanding and rearranging into a quadratic form ax^2 + Bx + C = 0, the coefficient of x must be zero for the sum of roots to be zero.

Multiple choice

Consider the quadratic equation $ax^{2} + bx + (a^{2} + b^{2} + c^{2} - ab - bc - ca) = 0$ (where $a,b, c$ are distinct real numbers) If the equation does not have any real roots then which of the following is always true.
  1. $b > 0$
  2. $b < 0$
  3. $a > 0$
  4. $a < 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The discriminant D = b^2 - 4a(a^2 + b^2 + c^2 - ab - bc - ca). For no real roots, D < 0. D = b^2 - 4a^3 - 4ab^2 - 4ac^2 + 4a^2b + 4abc + 4a^2c. This expression is complex, but for a quadratic ax^2+bx+k=0 to have no real roots, if a > 0, the vertex must be above the x-axis.

Multiple choice

Let $ax^2+a^2x+2=0$ be a quadratic equation, $a \in R$ and $S$ be the set of values of $a$ for which roots of this equation are imaginary or equal, then answer the following questions: Number of integral values of $a$ in set $S$ are
  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For ax^2 + a^2x + 2 = 0, roots are imaginary or equal if discriminant D <= 0. D = (a^2)^2 - 4(a)(2) = a^4 - 8a <= 0. a(a^3 - 8) <= 0. a(a-2)(a^2+2a+4) <= 0. Since a^2+2a+4 is always positive, we need a(a-2) <= 0, so 0 <= a <= 2. The integral values are 0, 1, 2. However, if a=0, the equation is not quadratic (2=0), so a cannot be 0. Thus, integral values are 1 and 2.

Multiple choice

If one root of the equation $ax^2 + bx + c = 0;&gt; a,b,c \in R$ is cube of other, then $\displaystyle \frac{b^2-2ac}{a+c}$ is equal to
  1. $\pm \sqrt{ac}$
  2. $\pm \sqrt{bc}$
  3. $\pm \dfrac{c^2}{a}$
  4. $\pm \dfrac{c^2}{b}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let roots be alpha and alpha^3. Sum = -b/a = alpha + alpha^3. Product = c/a = alpha^4. Then alpha = (c/a)^(1/4). Substitute into sum equation to relate a, b, c. The expression (b^2 - 2ac)/(a+c) simplifies to +/- sqrt(ac).

Multiple choice

If $\alpha$, $\beta$ are the roots of the equation $a{ x }^{ 2 }+bx+x=0$, then the roots of the equation $\left( a+b+c \right) { x }^{ 2 }-\left( b+2c \right) x+c=0$ are
  1. $\dfrac { \alpha +1 }{ \alpha } ,\dfrac { \beta +1 }{ \beta } $
  2. $\dfrac { \alpha -1 }{ \alpha } ,\dfrac { \beta -1 }{ \beta } $
  3. $\dfrac { \alpha }{ \alpha +1 } ,\dfrac { \beta }{ \beta +1 } $
  4. $\dfrac { \alpha }{ \alpha -1 } ,\dfrac { \beta }{ \beta -1 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice

If a,b,c >0 and $a=2b+3c$, then the roots of the equation $ax^2+bx+c=0$ are real if
  1. $\mid{\dfrac{a}{c}-11}\mid\geq4\sqrt{7}$
  2. $\mid{\dfrac{c}{a}-11}\mid>3\sqrt{7}$
  3. $\mid{\dfrac{b}{c}-4}\mid\geq4\sqrt{7}$
  4. $\mid{\dfrac{c}{b}-4}\mid\geq2\sqrt{7}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For real roots, the discriminant D = b^2 - 4ac >= 0. Substituting a = 2b + 3c into the inequality and rearranging leads to the condition involving the ratio of coefficients.