Quantitative Aptitude

Algebra

3,715 Questions
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Algebra Questions

Multiple choice

The equation $\left| 1-x \right| +{ x }^{ 2 }=5$ has

  1. A rational root and an irrational root

  2. Two rational

  3. Two irrational roots

  4. No real roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Case 1: x >= 1, (x-1) + x^2 = 5 => x^2 + x - 6 = 0 => (x+3)(x-2) = 0. x=2 is a valid root. Case 2: x < 1, (1-x) + x^2 = 5 => x^2 - x - 4 = 0. Using quadratic formula, x = (1 +/- sqrt(1 + 16))/2 = (1 +/- sqrt(17))/2. Since sqrt(17) is irrational, these are irrational roots. Thus, we have one rational root (2) and irrational roots.

Multiple choice

Each of the equations $3x^2 - 2 = 25, (2x - 1)^2 = (x - 1)^2, \sqrt{x^2 - 7} = \sqrt{x - 1}$ has

  1. two integral roots
  2. no root greater than $3$

  3. no root zero

  4. only one root

  5. only negative root and one positive root
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

1) 3x^2 = 27 -> x^2 = 9 -> x = 3, -3. 2) (2x-1)^2 = (x-1)^2 -> 2x-1 = x-1 (x=0) or 2x-1 = -(x-1) (3x=2, x=2/3). 3) sqrt(x^2-7) = sqrt(x-1) -> x^2-7 = x-1 -> x^2-x-6=0 -> (x-3)(x+2)=0. x=3 (valid), x=-2 (invalid). Roots are 3, -3, 0, 2/3, 3. None are greater than 3.

Multiple choice

The value of $a$ for which one root of the quadratic equation $(a^2-5a+3) x^2+(3a-1)x+2=0 $ is twice as large as the other, is :
  1. $-\cfrac{2}{3}$
  2. $\cfrac{1}{3}$
  3. $-\cfrac{1}{3}$
  4. $\cfrac{2}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let roots be alpha and 2*alpha. Sum of roots = -b/a = -(3a-1)/(a^2-5a+3) = 3*alpha. Product of roots = c/a = 2/(a^2-5a+3) = 2*alpha^2. From product, alpha^2 = 1/(a^2-5a+3). From sum, alpha = -(3a-1)/(3*(a^2-5a+3)). Squaring alpha: (3a-1)^2 / (9*(a^2-5a+3)^2) = 1/(a^2-5a+3). (9a^2 - 6a + 1) = 9*(a^2-5a+3) = 9a^2 - 45a + 27. 39a = 26. a = 26/39 = 2/3.

Multiple choice

Solve the following quadratic equation by factorization, $3\sqrt{5}x^2 \, + \, 25x \, - \, 10\sqrt{5} \, = \, 0$ the roots are
  1. $\dfrac{\sqrt5}{2}$ $ , -2\sqrt{5}$
  2. $\dfrac{\sqrt5}{3}$ $ , -2\sqrt{5}$
  3. $\dfrac{\sqrt5}{5}$ $ , -2\sqrt{5}$
  4. $\dfrac{\sqrt5}{2}$ $ , -2\sqrt{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

3*sqrt(5)x^2 + 25x - 10*sqrt(5) = 0. Multiply by sqrt(5): 15x^2 + 25*sqrt(5)x - 50 = 0. Divide by 5: 3x^2 + 5*sqrt(5)x - 10 = 0. Using quadratic formula: x = (-5*sqrt(5) +/- sqrt(125 - 4*3*(-10))) / 6 = (-5*sqrt(5) +/- sqrt(245)) / 6 = (-5*sqrt(5) +/- 7*sqrt(5)) / 6. Roots are 2*sqrt(5)/6 = sqrt(5)/3 and -12*sqrt(5)/6 = -2*sqrt(5).

Multiple choice

On Solving the following quadratic equation by factorization, the roots are $\dfrac{a^2 \, + \, b^2}{3}, \, \dfrac{b^2 \, - \, a^2}{3}$ : $9x^2 \, - \, 6b^2x \, - \, (a^4 \,- \, b^4) \, = \, 0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is 9x^2 - 6b^2x - (a^4 - b^4) = 0. Using the quadratic formula, x = [6b^2 +/- sqrt(36b^4 + 36(a^4 - b^4))] / 18 = [6b^2 +/- sqrt(36a^4)] / 18 = [6b^2 +/- 6a^2] / 18 = (b^2 +/- a^2) / 3. The roots are (b^2+a^2)/3 and (b^2-a^2)/3.

Multiple choice

Solve the following quadratic equation by factorization, the roots are : $48x^2$$ -$ $13x - 1 = 0$
  1. $-\dfrac{1}{16}, \, \dfrac{1}{2}$
  2. $-\dfrac{1}{16}, \, \dfrac{1}{4}$
  3. $-\dfrac{1}{6}, \, \dfrac{1}{3}$
  4. $-\dfrac{1}{16}, \, \dfrac{1}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Factor 48x^2 - 13x - 1 = 0. We need two numbers that multiply to -48 and add to -13, which are -16 and 3. So, 48x^2 - 16x + 3x - 1 = 0, which factors to 16x(3x - 1) + 1(3x - 1) = 0. The roots are x = 1/3 and x = -1/16.