Quantitative Aptitude

Algebra

3,715 Questions
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Algebra Questions

Multiple choice

If $a+b+c=0$ and a,b,c are rational, then the roots of the equation $(b+ c -a)x^2+(c+a-b)x+(a+b-c)=0$ are

  1. Rational

  2. Irrational

  3. Imaginary

  4. Equal
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a+b+c=0, then b+c = -a, c+a = -b, a+b = -c. The equation becomes (-a-a)x^2 + (-b-b)x + (-c-c) = 0 => -2ax^2 - 2bx - 2c = 0 => ax^2 + bx + c = 0. Since a, b, c are rational, the roots are rational if the discriminant b^2 - 4ac is a perfect square. However, the question implies general properties; given a+b+c=0, the roots are rational.

Multiple choice

The values of $k$, so that the equations $2x^{2} + kx - 5 = 0$ and $x^{2} - 3x - 4 = 0$ have one root in common, are
  1. $3, \dfrac {27}{2}$
  2. $9, \dfrac {27}{4}$
  3. $-3, \dfrac {-27}{4}$
  4. $-3, \dfrac {4}{27}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The roots of x^2 - 3x - 4 = 0 are (x-4)(x+1) = 0, so x = 4 or x = -1. If x = 4 is a common root, 2(4)^2 + k(4) - 5 = 0 => 32 + 4k - 5 = 0 => 4k = -27 => k = -27/4. If x = -1 is a common root, 2(-1)^2 + k(-1) - 5 = 0 => 2 - k - 5 = 0 => k = -3.

Multiple choice

Let us consider a quadratic equation $x^2+3ax+2a^2=0$ If this equation has roots $\alpha ,\beta $ and it is given that $\alpha^2 +\beta ^2=5$, then v alue of discriminant, $D$, for the above quadratic equation is
  1. $D>0$
  2. $D<0$
  3. $D=0$

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

For the roots, alpha^2 + beta^2 = (alpha + beta)^2 - 2alpha beta = 9a^2 - 4a^2 = 5a^2. Since this equals 5, a^2 = 1. The discriminant is 9a^2 - 8a^2 = a^2 = 1, so D > 0.

Multiple choice

Let us consider a quadratic equation $x^2+\lambda x+\lambda +1.25=0$, where $\lambda$ is a constant. The value of $\lambda$ such that the above quadratic equation has two coincident roots
  1. $\lambda =5$ or $\lambda=-1$
  2. $\lambda =1$ or $\lambda=5$
  3. $\lambda =-5$ or $\lambda=1$

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

For a quadratic equation to have coincident roots, the discriminant (D = b^2 - 4ac) must be zero. For x^2 + lambda*x + (lambda + 1.25) = 0, D = lambda^2 - 4(lambda + 1.25) = lambda^2 - 4*lambda - 5 = 0. Solving (lambda - 5)(lambda + 1) = 0 gives lambda = 5 or lambda = -1.

Multiple choice

The values of $y$ which will satisfy the equations $2x^{2} + 6x + 5y + 1 = 0, 2x + y + 3 = 0$ may be found by solving.
  1. $y^{2} + 14y - 7 = 0$
  2. $y^{2} + 8y + 1 = 0$
  3. $y^{2} + 10y - 7 = 0$
  4. $y^{2} + y - 12 = 0$

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From the second equation, x = (-y - 3) / 2. Substitute this into the first equation: 2((-y-3)/2)^2 + 6((-y-3)/2) + 5y + 1 = 0. This simplifies to 2((y^2 + 6y + 9)/4) - 3y - 9 + 5y + 1 = 0, which is (y^2 + 6y + 9)/2 + 2y - 8 = 0. Multiplying by 2 gives y^2 + 6y + 9 + 4y - 16 = 0, resulting in y^2 + 10y - 7 = 0.