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 equation whose roots are the squares of the roots of $x^{3}+ax+b=0$, is
  1. $x^{3}+2ax^{2}-a^{2}x-b^{2}=0$
  2. $x^{3}+2ax^{2}+a^{2}x-b^{2}=0$
  3. $x^{6}+ax^{2}+6=0$
  4. $x^{6}-ax^{2}+6=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let y = x^2, so x = sqrt(y). Substituting into x^3 + ax + b = 0 gives y*sqrt(y) + a*sqrt(y) + b = 0, or sqrt(y)(y + a) = -b. Squaring both sides: y(y + a)^2 = b^2, which is y(y^2 + 2ay + a^2) = b^2, or y^3 + 2ay^2 + a^2y - b^2 = 0.

Multiple choice

If $a, b, c$ are real distinct numbers satisfying the condition $a + b + c = 0$, then the roots of the quadratic equation $3ax^2 + 5bx + 7c = 0$ are

  1. positive

  2. negative

  3. real and distinct

  4. imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The discriminant D = (5b)^2 - 4(3a)(7c) = 25b^2 - 84ac. Since a+b+c=0, b = -(a+c). Substituting b gives 25(a+c)^2 - 84ac = 25a^2 + 50ac + 25c^2 - 84ac = 25a^2 - 34ac + 25c^2. Since a and c are real and distinct, this discriminant is always positive, ensuring real and distinct roots.

Multiple choice

If the roots of the equation $(a - 1) (x^2 + x + 1)^2 = (a + 1) (x^4 + x^2 + 1)$ are real and distinct, then the value of $a$ $\in$
  1. $(-\infty, 3 ]$
  2. $(- \infty, - 2) \cup (2, \infty)$
  3. $[-2, 2]$
  4. $[-3, \infty )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Dividing the equation by (x^2 + x + 1)^2 leads to (a - 1) = (a + 1) * ((x^4 + x^2 + 1) / (x^2 + x + 1)^2). Let t = (x^2 + 1) / x; the expression simplifies to a condition on the roots of a quadratic in t. Solving for real and distinct roots requires the discriminant to be positive, leading to the specified interval.

Multiple choice

Let $(a_1, a_2, a_3, a_4, a_5)$ denote a rearrangement of $(3, -5, 7, 4, -9)$, then the equation $a_1x^4+a_2x^3+a_3x^2+a_4x+a_5=0$ has

  1. At least two real roots.

  2. All four real roots.

  3. Only imaginary roots.

  4. Two real and two imaginary roots.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By Descartes' Rule of Signs, the number of sign changes in the coefficients determines the number of positive/negative roots. Given the set of coefficients, there are sign changes, ensuring at least two real roots.

Multiple choice

If $P(x)=ax^2+bx+c$ and $Q(x)=-ax^2+dx+c$ , $ac\neq 0$, then the equation $P(x). Q(x)=0$ has

  1. two imaginary roots

  2. more than two imaginary roots

  3. atleast two real roots

  4. no real roots
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

P(x)Q(x) = -(ax^2+bx+c)(ax^2-dx-c) = 0. This is a product of two quadratics. Since the leading coefficients of the two quadratics have opposite signs (a and -a), the product will have at least two real roots because the function will cross the x-axis.

Multiple choice

If $p, q, r$ & $s$ are real numbers & $pr = 2\left( q + s \right)$, then which of the following statement is correct about these equations ${ x }^{ 2 } + px + q = 0, { x }^{ 2 } + rx + s = 0$

  1. Both equations must have real roots

  2. Atleast one of the equation must have real roots

  3. Both equations must have imaginary roots

  4. Exactly one of the equation must have real roots
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Discriminants are D1 = p^2 - 4q and D2 = r^2 - 4s. Sum = D1 + D2 = p^2 + r^2 - 4(q + s). Given pr = 2(q + s), then 2pr = 4(q + s). So D1 + D2 = p^2 + r^2 - 2pr = (p - r)^2. Since the sum of discriminants is >= 0, at least one must be >= 0.

Multiple choice

The equation ${ x }^{ 2 } - ax + b = 0 & { x }^{ 3 } - p{ x }^{ 2 } + qx = 0$, where $b \neq 0, q \neq 0$, have one common root & the second equation has two equal roots. Then, $2\left( q + b \right) = $.
  1. $2ap$
  2. $ap^{2}$
  3. $a^2p$
  4. $ap$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the common root be k. k^2 - ak + b = 0 and k^3 - pk^2 + qk = 0. The second equation has two equal roots, say r, r, and s. So x^3 - (2r+s)x^2 + (r^2+2rs)x - r^2s = 0. Comparing coefficients, p = 2r+s, q = r^2+2rs. The common root k must be r. Substituting k=r into the first equation: r^2 - ar + b = 0, so b = ar - r^2. Then 2(q+b) = 2(r^2+2rs + ar - r^2) = 2(2rs + ar) = 2r(2s+a). Through substitution, this simplifies to ap.

Multiple choice

If the equation $x^2+b^2=1-2bx$ and $x^2+a^2=1-2ax$ have one and only one root common, then $|a-b|=$
  1. $1$
  2. $0$
  3. $2$

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

Subtracting the two equations: (x^2 + b^2 - 1 + 2bx) - (x^2 + a^2 - 1 + 2ax) = 0. b^2 - a^2 + 2bx - 2ax = 0. (b-a)(b+a) + 2x(b-a) = 0. Since a != b, divide by (b-a): (b+a) + 2x = 0, so x = -(a+b)/2. Substitute x back into one equation: (-(a+b)/2)^2 + b^2 = 1 - 2b(-(a+b)/2). (a+b)^2 / 4 + b^2 = 1 + b(a+b). (a^2 + 2ab + b^2) + 4b^2 = 4 + 4ab + 4b^2. a^2 - 2ab + b^2 = 4. (a-b)^2 = 4. |a-b| = 2.

Multiple choice

lf $\alpha,\ \beta$ are real and $\alpha^{2},-\beta^{2}$ are the roots of the equation $\mathrm{a}^{2}\mathrm{x}^{2}+\mathrm{x}+(1-\mathrm{a}^{2})=0(\mathrm{a}>1)$, then $\beta^{2}=$
  1. $ \mathrm{a}^{2}$
  2. $1$
  3. $1-\mathrm{a}^{2}$
  4. $1+\mathrm{a}^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the equation ax^2 + x + (1-a^2) = 0, the roots are alpha^2 and -beta^2. By Vieta's formulas, the product of roots is (1-a^2)/a^2. Thus, (alpha^2)(-beta^2) = (1-a^2)/a^2. Also, the sum of roots is -1/a^2. Given the structure, testing beta^2 = 1 yields roots that satisfy the equation.

Multiple choice

If $\alpha, \beta $ are the roots of equation $x^2-4x-1=0$, then the value of $\left ( \alpha ^{{1}/{3}}+\beta ^{{1}/{3}} \right)$ is
  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let S = alpha^(1/3) + beta^(1/3). S^3 = alpha + beta + 3 * (alpha*beta)^(1/3) * (alpha^(1/3) + beta^(1/3)). From x^2 - 4x - 1 = 0, alpha+beta = 4 and alpha*beta = -1. S^3 = 4 + 3 * (-1)^(1/3) * S. This leads to S^3 + 3S - 4 = 0. Testing S=1: 1 + 3 - 4 = 0. Thus, S=1 is a root.

Multiple choice

If the two equations $ x^{2}-cx+d=0 $ and $ x^{2}-ax+b=0 $ have a common root and the second equation has equal roots, then
  1. $b+d = ac$
  2. $2 (b+d) = ac$
  3. $b+d=2ac$
  4. $ (b+d)^{2} =a+c $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Eq 2: x^2 - ax + b = 0 has equal roots, so discriminant = a^2 - 4b = 0, meaning b = a^2/4. Let alpha be the common root. alpha^2 - c*alpha + d = 0 and alpha^2 - a*alpha + b = 0. Subtracting: (a-c)alpha + (d-b) = 0, so alpha = (b-d)/(a-c). Substituting back into x^2 - ax + b = 0: ((b-d)/(a-c))^2 - a((b-d)/(a-c)) + b = 0. This simplifies to 2(b+d) = ac.

Multiple choice

Statement I: lf $\alpha,\ \beta$ are the roots of $x^{2}-ax+b=0$, then the equation whose roots are $\displaystyle \frac{\alpha+\beta}{\alpha},\ \displaystyle \frac{\alpha+\beta}{\beta}$ is $bx^{2}-a^{2}x+a^{2}=0$ Statement II: lf $\alpha,\ \beta$ are the roots of $x^{2}-bx+c=0$ and $\alpha+h,\ \beta+h$ are the roots of $x^{2}+qx+r=0$, then $\mathrm{h}=\mathrm{b}-\mathrm{q}$. Which of the above statement(s) is(are) true.

  1. only I

  2. only II

  3. both I and II

  4. neither I and II
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Statement I: Roots are alpha, beta of x^2 - ax + b = 0. Sum = a, product = b. New roots are (a/alpha) and (a/beta). Sum = a(1/alpha + 1/beta) = a((alpha+beta)/alpha*beta) = a(a/b) = a^2/b. Product = a^2/(alpha*beta) = a^2/b. Equation: x^2 - (a^2/b)x + (a^2/b) = 0. Multiply by b: bx^2 - a^2x + a^2 = 0. Statement I is true. Statement II: Roots alpha, beta of x^2 - bx + c = 0. Sum = b, product = c. Roots alpha+h, beta+h of x^2 + qx + r = 0. Sum = alpha+beta+2h = -q. So b+2h = -q => 2h = -q-b. Statement II is false.