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

If $a$ and $b$ are positive real numbers and each of the equations $x^2+3ax+b=0$ and $x^2+bx+3a=0$ has real roots, then the smallest value of $(a+b)$ is
  1. $\frac{16}{3}$
  2. $6$
  3. $\frac{14}{3}$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the roots to be real, the discriminant must be non-negative. For x^2 + 3ax + b = 0, D = 9a^2 - 4b >= 0. For x^2 + bx + 3a = 0, D = b^2 - 12a >= 0. We want to minimize a + b. Testing values that satisfy both inequalities, the minimum occurs at a = 4/3 and b = 4, giving a + b = 16/3.

Multiple choice

Let $\alpha, \beta$ be the roots of the equation $x^2-x+p=0$ and $\gamma, \delta$ be the roots of $x^2-4x+q=0$. If $\alpha, \beta, \gamma, \delta$ are in G.P., then the integral values of p and q respectively, are?
  1. $-2, -32$
  2. $-2, 3$
  3. $-6, 3$
  4. $-6, -32$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Roots are in GP: a/r, a, ar, ar^2. Eq1: x^2-x+p=0 => sum=1, prod=p. Eq2: x^2-4x+q=0 => sum=4, prod=q. Let roots be a/r, a, ar, ar^2. Sums: a/r + a = 1 and ar + ar^2 = 4. a(1/r + 1) = 1 and ar(1+r) = 4. Divide: (ar(1+r)) / (a(1/r+1)) = 4/1 => r^2 = 4 => r=2 or -2. If r=2, a(1/2+1)=1 => a(3/2)=1 => a=2/3. Roots: 1/3, 2/3, 4/3, 8/3. p = (1/3)(2/3) = 2/9. q = (4/3)(8/3) = 32/9. The question asks for integral values, but these are not integers. Re-check: maybe roots are a, ar, ar^2, ar^3? Sums: a+ar=1, ar^2+ar^3=4 => a(1+r)=1, ar^2(1+r)=4 => r^2=4 => r=2 or -2. If r=-2, a(1-2)=1 => a=-1. Roots: -1, 2, -4, 8. p = (-1)*2 = -2. q = (-4)*8 = -32.

Multiple choice

If $ \alpha $ and $ \beta$ are the roots of the equation $ { ax }^{ 2 }+bx+c=0$ ($ a\neq 0,a.b,c$ being different), then $ \left( 1+\alpha +{ \alpha }^{ 2 } \right) \left( 1+\beta +{ \beta }^{ 2 } \right) $ is equal to

  1. Zero

  2. Positive

  3. Negative

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

The expression (1+alpha+alpha^2)(1+beta+beta^2) expands to 1 + (alpha+beta) + (alpha^2+beta^2) + alpha*beta + alpha^2*beta^2 + alpha*beta*(alpha+beta) + alpha^2*beta^2. Using Vieta's formulas (alpha+beta = -b/a, alpha*beta = c/a), this simplifies to a positive value for real roots of a quadratic.

Multiple choice

If the roots of the equation $\dfrac{x^{2}-bx}{ax-c}=\dfrac{m-1}{m+1}$ are equal but opposite in sign, then the value of $m$ will be
  1. $\dfrac{a-b}{a+b}$
  2. $\dfrac{ b-a}{a+b}$
  3. $\dfrac{a+b}{a-b}$
  4. $\dfrac{b+a}{b-a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the roots to be equal but opposite in sign, the sum of the roots must be zero. Rearranging the equation to standard quadratic form ax^2 + bx + c = 0, the coefficient of the linear term must be zero, which leads to the ratio m = (a-b)/(a+b).

Multiple choice

If ${x}^{2}+px+q=0$ is the quadratic equation whose roots are $a-2$ and $b-2$, where a and b are the roots of ${x}^{2}-3x+1=0$
  1. $p=1q=5$
  2. $p=1q=-5$
  3. $p=1q=-1$
  4. $none of these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots of x^2-3x+1=0 are a, b. Sum a+b=3, product ab=1. New roots are a-2, b-2. New sum = (a+b)-4 = 3-4 = -1. New product = (a-2)(b-2) = ab - 2(a+b) + 4 = 1 - 2(3) + 4 = -1. Equation is x^2 - (-1)x + (-1) = 0, so x^2 + x - 1 = 0. Thus p=1, q=-1.

Multiple choice

If the product of the roots of the equation $x^2-3kx+2e^{2:log:k}-1=0$ is $7$, then the roots of the equation are real if $k$ equals-
  1. $1$
  2. $3$
  3. $-4$
  4. $\pm2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Product of roots = c/a = 2e^(2 log k) - 1 = 7. Since e^(log k^2) = k^2, we have 2k^2 - 1 = 7, so 2k^2 = 8, k^2 = 4, k = +/- 2. For real roots, discriminant D = (-3k)^2 - 4(1)(7) >= 0, so 9k^2 - 28 >= 0. With k^2 = 4, 9(4) - 28 = 36 - 28 = 8 > 0, so roots are real.

Multiple choice

If $e^{\lambda}$ and $e^{-\lambda}$ are the roots of equation $3x^{2}-(a+b)x+2a=0,a,b,\lambda \epsilon R, \lambda \neq 0$ then least integral value of $b$ is
  1. $4$
  2. $5$
  3. $9$
  4. $10$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given roots e^lambda and e^-lambda, their product is e^lambda * e^-lambda = 1. From the quadratic equation 3x^2 - (a+b)x + 2a = 0, the product of roots is 2a/3. Thus, 2a/3 = 1, so a = 1.5. The sum of roots is e^lambda + e^-lambda = (a+b)/3. Since e^lambda + e^-lambda >= 2 for real lambda, (1.5+b)/3 >= 2, so 1.5+b >= 6, b >= 4.5. The least integral value of b is 5.

Multiple choice

If the roots of the equation $(b - c){x^2} + (c - a)x + (a - b) = 0$ be equal then b is equal to
  1. $\displaystyle {{a + c} \over 2}$
  2. $\displaystyle {{a - c} \over 2}$
  3. $\displaystyle{{a + c} \over {2ac}}$
  4. $\displaystyle{{a - c} \over {2ac}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a quadratic equation Ax^2 + Bx + C = 0 to have equal roots, the discriminant B^2 - 4AC must be zero. Here, (c-a)^2 - 4(b-c)(a-b) = 0. Expanding this leads to (a+c-2b)^2 = 0, so a+c = 2b, or b = (a+c)/2.

Multiple choice

If roots of the quadratic equation $3 a x ^ { 2 } + 2 b x + c = 0$ are in the ratio $2 : 3$, then which of the following statement is true?
  1. $8 a c = 25 b$
  2. $8 a c = 9 b ^ { 2 }$
  3. $8 b ^ { 2 } = 9 a c$
  4. $8 b ^ { 2 } = 25 a c$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the roots be 2k and 3k. Sum of roots = 5k = -2b/(3a), so k = -2b/(15a). Product of roots = 6k^2 = c/(3a). Substituting k gives 6 * (4b^2 / 225a^2) = c/(3a), which simplifies to 8b^2 = 25ac.