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 one root of the quadratic equation $2x^2+kx-6=0$ is $2$, find the value of k. Also, find the other root.
  1. $k=-1$   and   the other root is $=\dfrac{3}{2}$
  2. $k=1$   and   the other root is $=\dfrac{-3}{2}$
  3. $k=1$   and   the other root is $=\dfrac{3}{2}$
  4. $k=-1$   and   the other root is $=\dfrac{-3}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If x=2 is a root, then 2(2^2) + k(2) - 6 = 0, so 8 + 2k - 6 = 0, which means 2k = -2, so k = -1. The equation becomes 2x^2 - x - 6 = 0. Factoring: (2x + 3)(x - 2) = 0. The roots are 2 and -3/2.

Multiple choice

The number of quadratic equation $x^{2}+bx+c=0$ that have real and equal roots where coefficients $b$ and $c$ are from the set $\left{1,2,3,4,5\right}$ ($b$ and $c$ may be equal) is
  1. $4$
  2. $12$
  3. $2$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For real and equal roots, discriminant D = b^2 - 4c = 0, so b^2 = 4c. Given b, c in {1, 2, 3, 4, 5}: If b=1, c=1/4 (no). If b=2, c=1 (yes). If b=3, c=9/4 (no). If b=4, c=4 (yes). If b=5, c=25/4 (no). Only (2,1) and (4,4) work.

Multiple choice

If $\alpha$ and $\beta(\alpha > \beta)$ are roots of the equation $x^{2}-\sqrt{2}x+\sqrt{3-2\sqrt{2}}=0$, then the value of $(\cos^{-1}\alpha+\tan^{-1}\alpha+\tan^{-1}\beta)$ is equal to
  1. $\dfrac{3\pi}{8}$
  2. $\dfrac{5\pi}{8}$
  3. $\dfrac{7\pi}{8}$
  4. $\dfrac{\pi}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is x^2 - sqrt(2)x + (sqrt(3)-sqrt(2)) = 0? No, sqrt(3-2sqrt(2)) = sqrt((sqrt(2)-1)^2) = sqrt(2)-1. So x^2 - sqrt(2)x + sqrt(2)-1 = 0. Roots are 1 and sqrt(2)-1. Since alpha > beta, alpha = 1, beta = sqrt(2)-1. cos^-1(1) = 0. tan^-1(1) = pi/4. tan^-1(sqrt(2)-1) = pi/8. Sum = 0 + pi/4 + pi/8 = 3pi/8.

Multiple choice

If the roots $\alpha, \beta$ of the equation $px^{2} + qx + r = 0$ are real and of opposite sign (where $p, q, r$ are real coefficient), then the roots of the equation $\alpha (x - \beta)^{2} + \beta (x - \alpha)^{2} = 0$ are

  1. Positive

  2. Negative

  3. Real and of opposite sign

  4. Imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given roots alpha and beta of px^2 + qx + r = 0 are real and opposite sign. The equation alpha(x-beta)^2 + beta(x-alpha)^2 = 0 expands to (alpha+beta)x^2 - 4*alpha*beta*x + alpha*beta(alpha+beta) = 0. The discriminant D = 16*alpha^2*beta^2 - 4(alpha+beta)^2*alpha*beta. Since alpha and beta have opposite signs, alpha*beta < 0. Thus, D > 0, meaning roots are real. The product of roots is alpha*beta(alpha+beta)/(alpha+beta) = alpha*beta < 0, so roots are of opposite sign.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $2x^{2}-7x+8=0$, then the equation whose roots are $(3\alpha-4\beta$) and ($3\beta-4\alpha$) is
  1. $2x^{2}+7x+98=0$
  2. $x^{2}+7x+98=0$
  3. $2x^{2}-7x-98=0$
  4. $2x^{2}-7x+98=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given roots alpha, beta of 2x^2 - 7x + 8 = 0, alpha+beta = 3.5, alpha*beta = 4. Let new roots be y1 = 3a-4b, y2 = 3b-4a. Sum = -a-b = -3.5. Product = (3a-4b)(3b-4a) = 9ab - 12a^2 - 12b^2 + 16ab = 25ab - 12(a^2+b^2) = 25ab - 12((a+b)^2 - 2ab) = 25(4) - 12(12.25 - 8) = 100 - 12(4.25) = 100 - 51 = 49. Equation: x^2 - (sum)x + product = x^2 + 3.5x + 49 = 0, or 2x^2 + 7x + 98 = 0.