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 the roots of the equation, $x^{3}+3a\mathrm{x}^{2}+3bx+c=0$ are in H.P. then
  1. $b^{2}=c(3ab-c)$
  2. $2b^{3}=c(3ab-c)$
  3. $2b^{3}=c^{2} (3 {\it ab-c})$
  4. $2b^{2}=c^{2} (3 {\it ab-c})$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If roots of x^3 + 3ax^2 + 3bx + c = 0 are in HP, their reciprocals are in AP. Let roots be 1/p, 1/q, 1/r. The equation for reciprocals is cx^3 + 3bx^2 + 3ax + 1 = 0. For roots in AP, 2 * (second term coefficient / first term coefficient) = ... leads to the condition 2b^3 = c(3ab - c).

Multiple choice

The equations $\displaystyle 4x^{2}-x-1= 0$ and $\displaystyle 3x^{2}+\left ( \lambda +\mu \right )x+\lambda-\mu = 0$ have at least one common root , where $\lambda$ and $\mu$ are rational numbers. Find the value of $\lambda$ and $\mu$.
  1. $\displaystyle \lambda = 0, \mu =-\frac{3}{4}$
  2. $\displaystyle \lambda = -\frac{3}{4}, \mu = 0$
  3. $\displaystyle \lambda = -\frac{3}{4}, \mu = \frac{3}{4}$
  4. $\displaystyle \lambda = -\frac{3}{4}, \mu = \frac{1}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The roots of 4x^2 - x - 1 = 0 are x = (1 +/- sqrt(1 + 16))/8 = (1 +/- sqrt(17))/8. For the second equation to share a root, substituting x = (1 + sqrt(17))/8 into 3x^2 + (lambda + mu)x + (lambda - mu) = 0 must hold. Since lambda and mu are rational, the irrational parts must cancel, leading to lambda = -3/4 and mu = 0.

Multiple choice

Let $S$ be the set of values of $a$ for which $2$ lies between the roots of the quadratic equation $x^{2} + (a + 2)x - (a + 3) = 0$, then $S$ is given by
  1. $(-\infty , -5]$
  2. $(5 , \infty)$
  3. $(-\infty , -5)$
  4. $[5 , \infty)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For 2 to lie between the roots of x^2 + (a + 2)x - (a + 3) = 0, the value of the function f(x) = x^2 + (a + 2)x - (a + 3) at x = 2 must be negative. Substituting x = 2 gives 4 + 2(a + 2) - (a + 3) = 4 + 2a + 4 - a - 3 = a + 5. Setting a + 5 < 0 yields a < -5.

Multiple choice

If $a,b,c\in R$ and the quadratic equation $ax^{2}+bx+c=0$ has no real roots, then
  1. $(a + b + c) c > 0 $
  2. $c < 0$
  3. $a + b + c > 0$
  4. $a + b + c < 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a quadratic has no real roots, it is always positive (if a > 0) or always negative (if a < 0). Thus, f(x) = ax^2 + bx + c never changes sign. f(0) = c and f(1) = a + b + c must have the same sign. Therefore, their product c(a + b + c) must be positive.

Multiple choice

If $\displaystyle \alpha$, $\displaystyle \beta$, $\displaystyle \gamma$ are the roots of the equation $\displaystyle x^{3} + qx + r = 0$, then the equation whose roots are $\displaystyle \frac{\beta + \gamma}{\alpha^{2}}, : \frac{\gamma + \alpha}{\beta^{2}}, : \frac{\alpha + \beta}{\gamma^{2}}$, is
  1. $\displaystyle rx^{3} + qx^{2} + 1 = 0$
  2. $\displaystyle rx^{3} - qx^{2} - 1 = 0$
  3. $\displaystyle qx^{3} + rx^{2} + 1 = 0$
  4. $\displaystyle qx^{3} - rx^{2} - 1 = 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a standard transformation of roots for the cubic equation x^3 + qx + r = 0. The transformation leads to rx^3 - qx^2 - 1 = 0.

Multiple choice

If the equation $|x^2+bx+c|=k$ has four real roots, then
  1. $b^2-4c > 0$ and $0 < k < \dfrac {4c-b^2}{4}$
  2. $b^2-4c < 0$ and $0 < k < \dfrac {4c-b^2}{4}$
  3. $b^2-4c > 0$ and $k > \dfrac {4c-b^2}{4}$

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

The equation |x^2+bx+c|=k represents the intersection of the parabola y=x^2+bx+c and the horizontal lines y=k and y=-k. For four real roots, the vertex of the parabola must be below the x-axis (b^2-4c > 0) and the value of k must be between 0 and the absolute value of the y-coordinate of the vertex, which is (4c-b^2)/4.

Multiple choice

lf $a>0,b>0$, then roots of the equation $a-bx-x^{2}=0$ are:

  1. both positive

  2. both negative

  3. of opposite sign and numerically greater root is positive

  4. of opposite sign and numerically greater root is negative
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation is -x^2 - bx + a = 0, or x^2 + bx - a = 0. The product of roots is -a. Since a > 0, the product is negative, meaning the roots have opposite signs. The sum of roots is -b. Since b > 0, the sum is negative, meaning the root with the larger absolute value must be negative.

Multiple choice

Given $z = cos \displaystyle \left ( \frac{2\pi}{2n + 1} \right ) + i sin \left ( \frac{2\pi}{2n + 1} \right )$, n a positive integer, find the equation whose roots are $\alpha = z+ z^3 + ...... + z^{2n - 1}$ and $\beta = z^2 + z^4 + ..... + z^{2n}$
  1. $z^2 -z + \displaystyle \frac{1}{4} sec^2 \left ( \frac{\pi}{2n + 1} \right )$
  2. $z^2 +z + \displaystyle \frac{1}{4} sec^2 \left ( \frac{\pi}{2n + 1} \right )$
  3. $z^2 +z + \displaystyle \frac{1}{4} sec^2 \left ( \frac{\pi}{2n } \right )$
  4. $z^2 +z + \displaystyle \frac{1}{2} sec^2 \left ( \frac{\pi}{2n +1} \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a complex roots of unity problem. Given the structure of alpha and beta as sums of powers of z, they are roots of a quadratic equation derived from the properties of trigonometric sums. The correct form is z^2 + z + 1/4 * sec^2(pi / (2n+1)).

Multiple choice

If two sides of a triangle are the roots of the equation $4x^2 - (2 \sqrt{6}) x + 1= 0$ and the included angle is $60^o$, then the third side is
  1. $\sqrt{3}$
  2. $\dfrac{\sqrt{3}}{2}$
  3. $\dfrac1{\sqrt{3}}$
  4. $2\sqrt{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the roots be a and b. From the equation, a+b = (2*sqrt(6))/4 = sqrt(6)/2 and ab = 1/4. Using the Law of Cosines, c^2 = a^2 + b^2 - 2ab*cos(60). Since a^2 + b^2 = (a+b)^2 - 2ab = (6/4) - 2(1/4) = 1, c^2 = 1 - 2(1/4)*(1/2) = 1 - 0.25 = 0.75 = 3/4. Thus c = sqrt(3)/2.

Multiple choice

lf each root of the equation $3\mathrm{x}^{2} -7x +4=0$ is increased by $2$, then the resulting equation is
  1. $3\mathrm{x}^{2}-19\mathrm{x}+30=0$
  2. $3\mathrm{x}^{2}+5\mathrm{x}+2=0$
  3. $3\mathrm{x}^{2}-19\mathrm{x}+2=0$
  4. $3\mathrm{x}^{2}-19\mathrm{x}+20=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots are increased by 2, replace x with (x-2). 3(x-2)^2 - 7(x-2) + 4 = 0. 3(x^2 - 4x + 4) - 7x + 14 + 4 = 0. 3x^2 - 12x + 12 - 7x + 18 = 0. 3x^2 - 19x + 30 = 0.

Multiple choice

If the roots of the equation $px^2+2qx+r=0$ and $qx^2-2\sqrt{pr}x+q=0$ be real, then
  1. $p = q$
  2. $q^2=pr$
  3. $p^2=qr$
  4. $r^2=pq$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the roots to be real, the discriminant must be non-negative. For px^2 + 2qx + r = 0, D1 = (2q)^2 - 4pr = 4(q^2 - pr) >= 0, so q^2 >= pr. For qx^2 - 2sqrt(pr)x + q = 0, D2 = (-2sqrt(pr))^2 - 4(q)(q) = 4pr - 4q^2 >= 0, so pr >= q^2. Since q^2 >= pr and pr >= q^2, it must be that q^2 = pr.

Multiple choice

If the equations $\displaystyle 2x^{2}-7x+1=0$ and $ax^{2}+bx+2=0$ ($a ,b $ are rational) have a common root, then
  1. $a=2,b=-7$
  2. $\displaystyle a=-\frac{7}{2},b=1$
  3. $a=4,b=-14$

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

If they have a common root alpha, then 2*alpha^2 - 7*alpha + 1 = 0 and a*alpha^2 + b*alpha + 2 = 0. Multiplying the first by 2 gives 4*alpha^2 - 14*alpha + 2 = 0. Comparing coefficients with the second equation, we get a=4 and b=-14.