Quantitative Aptitude

Algebra

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Algebra Questions

Multiple choice

Number of rational roots of the equation $|x^{2}-2x -3|+4x =0$ is
  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Case 1: x^2 - 2x - 3 >= 0. (x-3)(x+1) >= 0, so x >= 3 or x <= -1. Equation: x^2 - 2x - 3 + 4x = 0 => x^2 + 2x - 3 = 0 => (x+3)(x-1) = 0. x = -3 (valid) or x = 1 (invalid). Case 2: x^2 - 2x - 3 < 0. -x^2 + 2x + 3 + 4x = 0 => -x^2 + 6x + 3 = 0 => x^2 - 6x - 3 = 0. Roots are (6 +/- sqrt(36 + 12))/2 = 3 +/- sqrt(12) = 3 +/- 2*sqrt(3). Neither is rational. Only x = -3 is a rational root.

Multiple choice

If the roots of the equation $(a+1)x^2-3ax+4a=0$ $(a\neq -1)$ are greater than unity, then the values of $a$ are.
  1. $\left[-\displaystyle\frac{16}{7}, -1\right)$
  2. $[0, -1]$
  3. $\displaystyle\left[-\frac{16}{7}, 1\right]$
  4. $[0, 1]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For roots of (a+1)x^2 - 3ax + 4a = 0 to be > 1: Let f(x) = (a+1)x^2 - 3ax + 4a. Conditions: D >= 0, (a+1)f(1) > 0, and -b/(2a) > 1. D = 9a^2 - 16a(a+1) = 9a^2 - 16a^2 - 16a = -7a^2 - 16a >= 0. This implies a(7a+16) <= 0, so a is in [-16/7, 0]. Checking f(1) = a+1 - 3a + 4a = 2a+1. (a+1)(2a+1) > 0. Vertex = 3a/(2(a+1)) > 1. Solving these yields the interval [-16/7, -1).

Multiple choice

Given $a, b, c$ are three distinct real numbers satisfying the inequality $a - 2b + 4c > 0$ and the equation $ax^{2} + bx + c = 0$ has no real roots. Then the possible value of $\dfrac {4a + 2b + c}{a + 3b + 9c}$ is/ are
  1. $2$
  2. $-1$
  3. $3$
  4. $\sqrt {2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The quadratic ax^2 + bx + c = 0 has no real roots, so the discriminant D = b^2 - 4ac < 0. The expression (4a + 2b + c)/(a + 3b + 9c) is evaluated at x=2 and x=3 respectively. Since the quadratic is always positive or negative, the ratio of values at different points relates to the function's properties. For this specific form, the value is 2.

Multiple choice

It is given that expression $a x ^ { 2 } + b x + c$ take negatives values for all $x < 7$.Then
  1. The equation $a x ^ { 2 } + b x + c = 0$,has equal roots

  2. a is negative:

  3. a and b is both negative:

  4. none of the foregoing statements correct'
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If a quadratic ax^2 + bx + c is negative for all x < 7, the parabola must open downwards (a < 0) and the root must be at x = 7. If it were positive for x < 7, a would be positive.

Multiple choice

Let $f(x)=4x^2-4ax+a^2-2a+2$ be a quadratic polynomial in $x$, $a \in R$. If at least one root of $f(x)=0$ lies in $[0,2]$, then the value of $a$ belongs too:
  1. $[1,5-\sqrt 7]$
  2. $[1,5+\sqrt 7]$
  3. $[\sqrt 7-5,\sqrt 7+5]$
  4. $(\sqrt 7-5,\infty)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The roots of 4x^2 - 4ax + a^2 - 2a + 2 = 0 are found using the quadratic formula: x = (4a +/- sqrt(16a^2 - 16(a^2 - 2a + 2))) / 8. This simplifies to x = (a +/- sqrt(8a - 8)) / 2. Setting these roots within [0, 2] leads to the interval [1, 5 + sqrt(7)].