Quantitative Aptitude

Algebra

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Algebra Questions

Multiple choice

If $\left( {2{x^2}\, - \,3x\, + \,1} \right)\left( {2{x^2}\, + \,5x\, + \,1} \right)\, = \,9{x^2}$, then equation has

  1. four real roots

  2. two real and two imaginary roots

  3. four imaginary roots

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

The equation (2x^2 - 3x + 1)(2x^2 + 5x + 1) = 9x^2 can be rewritten by dividing by x^2: (2x - 3 + 1/x)(2x + 5 + 1/x) = 9. Let y = 2x + 1/x. Then (y - 3)(y + 5) = 9, so y^2 + 2y - 15 = 9, or y^2 + 2y - 24 = 0. Factoring gives (y + 6)(y - 4) = 0. For y = 4, 2x + 1/x = 4 implies 2x^2 - 4x + 1 = 0 (discriminant 16 - 8 = 8 > 0, two real roots). For y = -6, 2x + 1/x = -6 implies 2x^2 + 6x + 1 = 0 (discriminant 36 - 8 = 28 > 0, two real roots). Total is four real roots.

Multiple choice

For what value of m, the roots of the equation $x^2-x+m=0$ are non real?
  1. $(\dfrac{1}{4}, \infty )$
  2. $(-\infty, \dfrac{1}{4})$
  3. $(-\dfrac{1}{4}, \dfrac{1}{4})$

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

For a quadratic equation ax^2 + bx + c = 0 to have non-real roots, the discriminant D = b^2 - 4ac must be less than 0. Here, (-1)^2 - 4(1)(m) < 0, which simplifies to 1 - 4m < 0, or m > 1/4.

Multiple choice

If there is a multiple root of order $3$ for the equation ${x}^{4}-2{x}^{3}+2x-a=0$, then the other root is
  1. $-1$
  2. $0$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let f(x) = x^4 - 2x^3 + 2x - a. If there is a root of order 3, then f(x) = (x-r)^3 * (x-k). Expanding this: (x^3 - 3x^2r + 3xr^2 - r^3)(x-k) = x^4 - (k+3r)x^3 + (3r^2 + 3rk)x^2 - (r^3 + 3r^2k)x + kr^3. Comparing coefficients with x^4 - 2x^3 + 0x^2 + 2x - a: k+3r = 2, 3r^2 + 3rk = 0, r^3 + 3r^2k = -2. From 3r^2 + 3rk = 0, r(r+k) = 0. If r=0, then k=2, but then the x term is 0, not 2. If r+k=0, k=-r. Then k+3r = 2 becomes -r+3r=2, so 2r=2, r=1. Then k=-1. The root of order 3 is 1, and the other root is -1.

Multiple choice

Equation $\sqrt{5x^{2}-6x+8}-\sqrt{5x^{2}-6x-7}=1$ has

  1. All roots positive

  2. All roots negative

  3. One root positive and the other negative

  4. Two root positive and one root negative
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let y = 5x^2 - 6x. The equation is sqrt(y+8) - sqrt(y-7) = 1. Squaring gives (y+8) + (y-7) - 2*sqrt((y+8)(y-7)) = 1, so 2y + 1 - 1 = 2*sqrt((y+8)(y-7)). y = sqrt(y^2 + y - 56), y^2 = y^2 + y - 56, y = 56. 5x^2 - 6x - 56 = 0. Roots are (6 +/- sqrt(36 - 4*5*(-56))) / 10 = (6 +/- sqrt(1156)) / 10 = (6 +/- 34) / 10. Roots are 4 and -2.8. One positive, one negative.