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

Solve the quadratic equation $$3x^2 + 7x - 6 = 0$$ using Brahmagupta's Formula.

  1. $$x = \frac{-7 \pm \sqrt{85}}{6}$$
  2. $$x = \frac{-7 \pm \sqrt{91}}{6}$$
  3. $$x = \frac{-7 \pm \sqrt{97}}{6}$$
  4. $$x = \frac{-7 \pm \sqrt{103}}{6}$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-7 \pm \sqrt{7^2 - 4(3)(-6)}}{2(3)}$$. Simplifying this, we get $$x = \frac{-7 \pm \sqrt{85}}{6}$$. Therefore, the solution set is ({\frac{-7 \pm \sqrt{85}}{6})).

Multiple choice

Solve the quadratic equation $$4x^2 - 12x + 9 = 0$$ using Brahmagupta's Formula.

  1. $$x = \frac{6 \pm \sqrt{15}}{2}$$
  2. $$x = \frac{6 \pm \sqrt{21}}{2}$$
  3. $$x = \frac{6 \pm \sqrt{27}}{2}$$
  4. $$x = \frac{6 \pm \sqrt{33}}{2}$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(4)(9)}}{2(4)}$$. Simplifying this, we get $$x = \frac{6 \pm \sqrt{15}}{2}$$. Therefore, the solution set is ({\frac{6 \pm \sqrt{15}}{2})).

Multiple choice

Solve the quadratic equation $$5x^2 + 2x - 3 = 0$$ using Brahmagupta's Formula.

  1. $$x = \frac{-1 \pm \sqrt{23}}{5}$$
  2. $$x = \frac{-1 \pm \sqrt{29}}{5}$$
  3. $$x = \frac{-1 \pm \sqrt{31}}{5}$$
  4. $$x = \frac{-1 \pm \sqrt{37}}{5}$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-2 \pm \sqrt{2^2 - 4(5)(-3)}}{2(5)}$$. Simplifying this, we get $$x = \frac{-1 \pm \sqrt{23}}{5}$$. Therefore, the solution set is ({\frac{-1 \pm \sqrt{23}}{5})).

Multiple choice

Solve the quadratic equation $$6x^2 - 11x + 3 = 0$$ using Brahmagupta's Formula.

  1. $$x = \frac{11 \pm \sqrt{109}}{12}$$
  2. $$x = \frac{11 \pm \sqrt{113}}{12}$$
  3. $$x = \frac{11 \pm \sqrt{127}}{12}$$
  4. $$x = \frac{11 \pm \sqrt{131}}{12}$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(6)(3)}}{2(6)}$$. Simplifying this, we get $$x = \frac{11 \pm \sqrt{109}}{12}$$. Therefore, the solution set is ({\frac{11 \pm \sqrt{109}}{12})).

Multiple choice

Solve the quadratic equation $$7x^2 - 13x + 6 = 0$$ using Brahmagupta's Formula.

  1. $$x = \frac{13 \pm \sqrt{121}}{14}$$
  2. $$x = \frac{13 \pm \sqrt{127}}{14}$$
  3. $$x = \frac{13 \pm \sqrt{133}}{14}$$
  4. $$x = \frac{13 \pm \sqrt{139}}{14}$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Brahmagupta's Formula, we have $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Substituting the values of a, b, and c, we get $$x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(7)(6)}}{2(7)}$$. Simplifying this, we get $$x = \frac{13 \pm \sqrt{121}}{14}$$. Therefore, the solution set is ({\frac{13 \pm \sqrt{121}}{14})).

Multiple choice

If one root of the equation ax2 + bx + c = 0 is twice the other, then

  1. 2b2 = 9ac

  2. b2 = 4ac

  3. b2 = 9ac

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

Let the roots be alpha and 2*alpha. The sum of roots is -b/a = 3*alpha, so alpha = -b/(3a). The product of roots is c/a = 2*alpha^2 = 2*(-b/(3a))^2 = 2*b^2/(9a^2). Simplifying gives c/a = 2*b^2/(9a^2), which leads to 9ac = 2b^2.

Multiple choice

Which of the following can be that possible equation whose roots are 10 times the roots of the given equation? 2x2 – 15x + 18 = 0

  1. x2 + 75x + 900 = 0

  2. x2 - 75x - 900 = 0

  3. x2 + 75x - 900 = 0

  4. x2 - 75x + 900 = 0

  5. Both (3) and (4)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If roots are 10 times the original roots, replace x with x/10 in the original equation: 2(x/10)^2 - 15(x/10) + 18 = 0. 2(x^2/100) - 1.5x + 18 = 0. Multiply by 50: x^2 - 75x + 900 = 0.

Multiple choice

If r and s are the solutions of quadratic equation mx2 + nx + c = 0, which of the following equations can have its roots equal to r2 and s2?

  1. m2x2 – n2x + 2mcx + c2 = 0

  2. n2x2 – m2x + 2mcx + c2 = 0

  3. m2x2 + n2x – 2mcx + c2 = 0

  4. mx2 – n2x + 2mcx + c2 = 0

  5. mx2 – nx + 2mcx + c = 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If r, s are roots of mx^2 + nx + c = 0, then r+s = -n/m and rs = c/m. We want an equation with roots r^2, s^2. Sum = r^2+s^2 = (r+s)^2 - 2rs = n^2/m^2 - 2c/m = (n^2 - 2mc)/m^2. Product = r^2s^2 = c^2/m^2. The equation is x^2 - (Sum)x + Product = 0, which is x^2 - ((n^2-2mc)/m^2)x + c^2/m^2 = 0. Multiplying by m^2 gives m^2x^2 - (n^2-2mc)x + c^2 = 0, or m^2x^2 - n^2x + 2mcx + c^2 = 0.