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 $\sec\alpha$ and cosec $\alpha$ are the roots of the equation $x^{2}-px+q=0$, then
  1. $p^{2}=q(q-2)$
  2. $p^{2}=q(q+2)$
  3. $p^{2}+q^{2}=2q$

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

Roots are sec(a) and cosec(a). Sum of roots = sec(a) + cosec(a) = p. Product of roots = sec(a) * cosec(a) = q. (sec(a) + cosec(a))^2 = p^2 => sec^2(a) + cosec^2(a) + 2*sec(a)*cosec(a) = p^2. Since sec^2(a) + cosec^2(a) = (sin^2(a) + cos^2(a)) / (sin^2(a)cos^2(a)) = 1 / (sin^2(a)cos^2(a)) = sec^2(a)cosec^2(a) = q^2, we have q^2 + 2q = p^2.

Multiple choice

A quadratic equation whose roots are $\displaystyle \tan 22\frac{1}{2}^{0}$ and $\displaystyle \cot 22 \frac{1}{2}^{0}$ is
  1. $x^{2}-2\sqrt{2}x+1=0$
  2. $x^{2_{-}}\sqrt{x}+1=0$
  3. $x^{2}+2\sqrt{2}x-1=0$
  4. $x^{2}-2\sqrt{2}-1=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let x1 = tan(22.5) and x2 = cot(22.5). Since cot(22.5) = 1/tan(22.5), the product of roots is 1. The sum of roots is tan(22.5) + cot(22.5) = (sin^2(22.5) + cos^2(22.5)) / (sin(22.5)cos(22.5)) = 1 / (0.5 * sin(45)) = 1 / (0.5 * 1/sqrt(2)) = 2*sqrt(2). The quadratic equation is x^2 - (sum)x + (product) = 0, which is x^2 - 2*sqrt(2)x + 1 = 0.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $a \cos \ 2 \theta + b \sin \ 2 \theta = c$, then $\tan \ ( \alpha + \beta )$ is equal to:
  1. $\dfrac{a}{b}$
  2. $\dfrac{-b}{a}$
  3. $\dfrac{b}{a}$
  4. $\dfrac{-a}{b}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation a cos(2theta) + b sin(2theta) = c can be written as a(1-tan^2(theta))/(1+tan^2(theta)) + b(2tan(theta))/(1+tan^2(theta)) = c. This is a quadratic in tan(theta). Let t = tan(theta). Then a(1-t^2) + 2bt = c(1+t^2), or (a+c)t^2 - 2bt + (c-a) = 0. The sum of roots tan(alpha)+tan(beta) = 2b/(a+c) and product tan(alpha)tan(beta) = (c-a)/(a+c). Using tan(alpha+beta) = (tan(alpha)+tan(beta))/(1-tan(alpha)tan(beta)), we get (2b/(a+c)) / (1 - (c-a)/(a+c)) = (2b/(a+c)) / (2a/(a+c)) = b/a.

Multiple choice

If $\displaystyle \alpha$ and $\displaystyle \beta$ are the roots of the equation a $\displaystyle \cos : 2 \theta + b : \sin : 2\theta = c$ then $\displaystyle \cos^2\alpha + \cos^2\beta$ is equal to
  1. $\displaystyle \frac {a^2 + ac + b^2}{a^2 + b^2}$
  2. $\displaystyle \frac {a^2 - ac + b^2}{a^2 + b^2}$
  3. $\displaystyle \frac {2b^2}{a^2 + c^2}$
  4. $\displaystyle \frac {2a^2}{b^2 + c^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given a cos(2theta) + b sin(2theta) = c, use the identities cos(2theta) = 2cos^2(theta) - 1 and sin(2theta) = 2sin(theta)cos(theta). This leads to a quadratic in tan(theta) or similar. The sum of cos^2(alpha) + cos^2(beta) for such equations is a standard result derived from the roots of the quadratic equation.

Multiple choice

If ABCD is a convex quadrilateral such that $\displaystyle 4\sec A+5=0$, then the quadratic equation whose roots are $\displaystyle \tan A$ and $\displaystyle co\sec A$ is
  1. $\displaystyle 12x^{2}-29x+15=0$
  2. $\displaystyle 12x^{2}-11x-15=0$
  3. $\displaystyle 12x^{2}+11x-15=0$

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

Given 4sec A + 5 = 0, sec A = -5/4. Since sec A is negative, A is in the second quadrant. Thus, tan A = -3/4 and cosec A = 5/3. The quadratic equation with roots -3/4 and 5/3 is (x + 3/4)(x - 5/3) = 0, which simplifies to x^2 - (5/3 - 3/4)x - 5/4 = 0, or x^2 - (11/12)x - 5/4 = 0. Multiplying by 12 gives 12x^2 - 11x - 15 = 0.

Multiple choice

For the equation $ax^2+bx+c=0,$ $a,b$ and $c$ are real, Statement 1: If the equation $ax^2+bx+c=0, 0 < a < b < c$, has non-real complex roots $z_1$ and $z_2$, then $|z_1| > 1, |z_2| > 1$. Statement 2: Complex roots always occur in conjugate pairs.

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

Statement 1: If $z_1, z_2$ are the roots of the quadratic equation $az^2+bz+c=0$ such that $Im(z_1z_2)\neq 0$, then at least one of a, b, c is imaginary. Statement 2: If quadratic equation having real coefficients has complex roots, then roots are always conjugate to each other.

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Statement 1 is true because if the coefficients are real, the roots must be real or conjugate pairs. If the roots are conjugate pairs (z, z_bar), their product is |z|^2, which is real, so Im(z1z2) would be 0. Thus, for the imaginary part to be non-zero, the coefficients cannot all be real. Statement 2 is a standard property of quadratic equations with real coefficients.

Multiple choice

The equation $x(x+2)(x^{2}-x)=-1$, has

  1. All roots imaginary

  2. All roots negative

  3. Two roots real and two roots imaginary

  4. All roots real
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

x(x+2)(x^2-x) = -1. x(x+2)x(x-1) = -1. x^2(x+2)(x-1) = -1. x^2(x^2 + x - 2) = -1. x^4 + x^3 - 2x^2 + 1 = 0. Testing for real roots: let f(x) = x^4 + x^3 - 2x^2 + 1. f'(x) = 4x^3 + 3x^2 - 4x = x(4x^2 + 3x - 4). Roots of f'(x) are 0 and (-3 +/- sqrt(9 + 64))/8 = (-3 +/- 8.5)/8. The function values at these points are all positive, meaning the minimum value is positive. Thus, there are no real roots.

Multiple choice

If 'a' is a complex number such that $|a| = 1$. Find the value of a, so that equation $az^2 + z + 1= 0$ has one purely imaginary root.
  1. $\alpha = cos^{-1} \displaystyle \left ( \frac{\sqrt{5} - 1}{2} \right )$
  2. $\alpha = cos^{-1} \displaystyle \left ( \frac{\sqrt{5} + 1}{2} \right )$
  3. $\alpha = cos^{-1} \displaystyle \left ( \frac{\sqrt{5}}{2} \right )$
  4. $\alpha = cos^{-1} \displaystyle \left ( \frac{\sqrt{3}}{2} \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer