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

The equations $x^{2} + x + a = 0$ and $x^{2} + ax + 1 = 0$ have a common real root
  1. For no value of $a$
  2. For exactly one value of $a$
  3. For exactly two values of $a$
  4. For exactly three values of $a$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Subtracting the equations: (x^2 + x + a) - (x^2 + ax + 1) = 0 => x(1-a) - (1-a) = 0 => (x-1)(1-a) = 0. Either x=1 or a=1. If x=1, 1+1+a=0 => a=-2. If a=1, x^2+x+1=0 (no real roots). Thus, only a=-2 works.

Multiple choice

If the ratio of the roots of the equation $p{ x }^{ 2 }+qx+r=0\quad $ is $a: b$, then $\cfrac { ab }{ { \left( a+b \right) }^{ 2 } } $ is equal to
  1. $\cfrac { { p }^{ 2 } }{ qr } $
  2. $\cfrac { pr }{ { q }^{ 2 } } $
  3. $\cfrac { { q }^{ 2 } }{ pr } $
  4. $\cfrac { pq }{ { r }^{ 2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For roots r1, r2 with ratio a:b, r1/r2 = a/b. Sum = -q/p, Product = r/p. (r1+r2)^2 / (r1*r2) = (-q/p)^2 / (r/p) = q^2 / pr. Also (r1+r2)^2 / (r1*r2) = (a+b)^2 / ab. Thus ab/(a+b)^2 = pr/q^2.

Multiple choice

If roots of equation ${ x }^{ 2 }+x+1=0$ are $a$, $b$ and roots of ${ x }^{ 2 }+px+q=0$ are $\dfrac { a }{ b } $, $\dfrac { b }{ a }$; then the value of $p+q$ is
  1. $2$
  2. $\dfrac { \sqrt { 2 } +1 }{ 2 } $
  3. $-1$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots of x^2+x+1=0 are omega and omega^2. The roots of the second equation are (a/b) and (b/a), which are (omega/omega^2) = omega^2 and (omega^2/omega) = omega. Thus, the second equation is also x^2+x+1=0, meaning p=1 and q=1. Therefore, p+q=2.

Multiple choice

Product of roots of the equation $\sqrt{13-x^2}=x+5$ is
  1. $-6$
  2. $7$
  3. $6$
  4. $-7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Squaring both sides of sqrt(13-x^2) = x+5 gives 13-x^2 = x^2+10x+25, which simplifies to 2x^2+10x+12 = 0 or x^2+5x+6 = 0. The roots are x = -2 and x = -3. Checking these in the original equation, sqrt(13-4) = 3 and -2+5 = 3 (valid), while sqrt(13-9) = 2 and -3+5 = 2 (valid). The product is (-2)*(-3) = 6.

Multiple choice

If the roots of the equation $ax^2+bx+c=0$ are real and of the form $\displaystyle\frac{\alpha}{\alpha -1}$ and $\displaystyle\frac{\alpha +1}{\alpha}$, then the value of $(a+b+c)^2$ is?
  1. $b^2-4ac$
  2. $b^2-2ac$
  3. $2b^2-ac$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots are r1 and r2, then r1+r2 = -b/a and r1*r2 = c/a. Sum = alpha/(alpha-1) + (alpha+1)/alpha = (alpha^2 + alpha^2 - 1) / (alpha^2 - alpha) = (2alpha^2 - 1) / (alpha^2 - alpha). Product = (alpha+1)/(alpha-1) = c/a. Algebraic manipulation leads to (a+b+c)^2 = b^2 - 4ac.

Multiple choice

If $6,8$ and $12$ are $l ^ { th }, m ^ { th }$ and $n ^ { th }$ terms of an A.P and $f(x)=nx^{ { 2 } }+2lx-2m,$ then the equation $f ( x ) = 0$ has
  1. a root between $0$ and $1$

  2. both roots imaginary

  3. both roots negative.
  4. both roots greater than $1.$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Terms of AP: a + (l-1)d = 6, a + (m-1)d = 8, a + (n-1)d = 12. f(x) = nx^2 + 2lx - 2m. Using the properties of AP terms, one can show f(0) = -2m and f(1) = n + 2l - 2m. Given the values, f(0) < 0 and f(1) > 0, implying a root in (0, 1).

Multiple choice

Solve the following equation. $2^x+2^y=20, x+y=6$.

  1. x=2,y=4

  2. x=3,y=3

  3. x=3,y=2

  4. x=0,y=2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given x+y=6, we can substitute y=6-x into the first equation to get 2^x + 2^(6-x) = 20. Multiplying by 2^x gives (2^x)^2 - 20(2^x) + 64 = 0, which factors to (2^x - 16)(2^x - 4) = 0. This yields 2^x=16 (x=4, y=2) or 2^x=4 (x=2, y=4).

Multiple choice

If $a>1$ is a real number, then the number of roots of the equation $a^{2\log_2x}$ = 5+ 4${x}^{\log_2a}$ has

  1. 2

  2. infinite

  3. 0

  4. 1
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let y = x^(log2 a). The equation becomes a^(2 log2 x) = y^2. Given a^(log2 x) = x^(log2 a), the equation simplifies to y^2 = 5 + 4y, or y^2 - 4y - 5 = 0. Factoring gives (y-5)(y+1) = 0, so y=5 or y=-1. Since y = x^(log2 a) must be positive, only y=5 is valid, yielding exactly one solution for x.

Multiple choice

If $\displaystyle a> 1$ is a real number, then the number of roots of the equation $\displaystyle a^{2\log_{2}x}= 5+4x^{\log_{2}a}$ is/are
  1. $2$

  2. infinite
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the logarithmic identity x^(log_b(a)) = a^(log_b(x)), we can substitute t = a^(log_2(x)) to rewrite the equation as t^2 - 4t - 5 = 0. Solving this quadratic equation gives t = 5 or t = -1. Since t = a^(log_2(x)) must be positive, we only have t = 5, which yields exactly one unique solution for x.