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 the roots of the equation $(\lambda+1)x^3-(\lambda^2+2)x^2-(\lambda^2+2)x +(\lambda+1)=0; \lambda \epsilon R$ are in a $A.P$. with non-zero common difference and one of the roots is independent of $\lambda$, then the sum of all possible values of $\lambda$ is $m$. Find the value of $4m$.
  1. $4$
  2. $5$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation is (lambda+1)x^3 - (lambda^2+2)x^2 - (lambda^2+2)x + (lambda+1) = 0. If x= -1, the equation becomes -(lambda+1) - (lambda^2+2) + (lambda^2+2) + (lambda+1) = 0. So x = -1 is a root independent of lambda. Let roots be a/r, a, ar. Product of roots = -d/a = -(lambda+1)/(lambda+1) = -1. So a^3 = -1, a = -1. Since -1 is a root, the other roots are -1/r and -r. Sum = -1 - 1/r - r = (lambda^2+2)/(lambda+1). This leads to a quadratic in lambda.

Multiple choice

Consider the following statements: I. The equation of lowest degree with rational coefficients, one of whose roots is $\sqrt{7}+\sqrt{3}$ is $x^{4}-20x^{2}+16=0$. II. The equation of lowest degree with rational coefficients, one of whose roots is $\sqrt{2}+i\sqrt{3}$ is $x^{4}+2x^{2}+25=0$.

  1. only I is true

  2. only II is true

  3. both I and II are true

  4. neither I nor II ture
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For I, if x = sqrt(7) + sqrt(3), then x^2 = 7 + 3 + 2*sqrt(21) = 10 + 2*sqrt(21). (x^2 - 10)^2 = 4*21 = 84, so x^4 - 20x^2 + 100 = 84, giving x^4 - 20x^2 + 16 = 0. For II, if x = sqrt(2) + i*sqrt(3), x^2 = 2 - 3 + 2*i*sqrt(6) = -1 + 2*i*sqrt(6). (x^2 + 1)^2 = -24, so x^4 + 2x^2 + 1 + 24 = 0, giving x^4 + 2x^2 + 25 = 0.

Multiple choice

The value of $\displaystyle a$ for which the equation $\displaystyle \left ( a^{2} - 3a + 2 \right )x^{2} + \left ( a^{2} - 5a + 4 \right )x + \left ( a^{2} - 1 \right ) = 0$ has more than two roots is:
  1. $\displaystyle -1$
  2. $\displaystyle 4$
  3. $\displaystyle 2$
  4. $\displaystyle 1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a quadratic to have more than two roots, it must be an identity (0=0). This requires coefficients of x^2, x, and the constant term to be zero. a^2-1=0 implies a=1 or a=-1. If a=1, the equation becomes 0x^2 + 0x + 0 = 0, which is true for all x.

Multiple choice

The equation $5x^2-2x+3$ has-

  1. Five distinct real roots

  2. Three distinct real roots and two complex roots

  3. One distinct real root and two complex roots

  4. One distinct real root and four complex roots

  5. None of these
Reveal answer Fill a bubble to check yourself
E Correct answer
Multiple choice

If the sum of the roots of the equation $a x ^ { 2 } + bx + c = 0$ is equal to the sum of the square of their reciprocal, then _________________.
  1. $c ^ { 2 } b , a ^ { 2 } c , b ^ { 2 } a$ are in A.P.
  2. $c ^ { 2 } b , a ^ { 2 } c , b ^ { 2 } a$ are in G.P.
  3. $\frac { b } { c } , \frac { a } { b } , \frac { c } { a }$ are in H.P.
  4. $\frac { b } { c } , \frac { a } { b } , \frac { c } { a }$ are in G.P.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let roots be p and q. p+q = -b/a and pq = c/a. Given p+q = 1/p^2 + 1/q^2 = (p^2+q^2)/(pq)^2 = ((p+q)^2 - 2pq)/(pq)^2. Substituting values: -b/a = (b^2/a^2 - 2c/a) / (c^2/a^2) = (b^2 - 2ac) / c^2. This simplifies to -bc^2 = a(b^2 - 2ac), or 2a^2c = ab^2 + bc^2. This implies bc^2, a^2c, b^2a are in A.P. as 2(a^2c) = bc^2 + ab^2.

Multiple choice

Let $\alpha,\beta$ be the roots of the equation $x^{2}-ax + b = 0$. If the coordinates of $A_{n}$ are $(\alpha^{n/2},\beta^{n/2})$ then,$(OA_{n+1})^{2} -a(OA_{n})^{2} + b(OA_{n-1})^{2}$ is equal to zero,O being the origin

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

  3. Assertion is correct but Reason is incorrect

  4. Both Assertion and Reason are incorrect
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

If OA and OB are two equal chords of the circle $x^2+y^2-2x+4y=0$ perpendicular to each other and passing through the origin, the slopes of OA and OB are the roots of the equation.
  1. $3m^2+8m-3=0$
  2. $3m^2-8m-3=0$
  3. $8m^2+3m-8=0$
  4. $8m^2+3m+8=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The circle x^2 + y^2 - 2x + 4y = 0 has center (1, -2). Chords OA and OB pass through origin (0,0). Since they are perpendicular, their slopes m1 and m2 satisfy m1 * m2 = -1. The equation of the lines is y = mx. Substituting into the circle equation: x^2 + (mx)^2 - 2x + 4(mx) = 0 -> x^2(1+m^2) - 2x(1-2m) = 0. For the chords, x(1+m^2) = 2(1-2m). This leads to the quadratic in m.

Multiple choice

The x-coordinates of the vertices of a square are the roots of the $x^{2} - 3|x| + 2 = 0$ and the y-coordinates of the vertices are the roots of the equation $y^{2} - 3y=0$ the possible vertices of the square is/ are
  1. $(1, 0), (2, 3), (2, 0), (1, 3)$
  2. $(-1, 3), (-2, 3), (-2, 0), (-1, 0)$
  3. $(2, 0), (1, 3), (-1, 0), (2, 0)$
  4. $(-2, 0), (-2, 3), (1, 0), (1, 3)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The roots of x^2 - 3|x| + 2 = 0 are x = 1, 2, -1, -2. The roots of y^2 - 3y = 0 are y = 0, 3. The vertices are formed by combinations of these coordinates. Option D correctly lists four points that form a square with side length 3.

Multiple choice

If for $0 < x< \pi/2$, $e^{[(\sin ^{2}x+\sin ^{4}x+\sin ^{6}x+...+\infty )\log_{e}2]}$ satisfies the quadratic equation, $x^{2}-9x+8=0$, find the value of $\displaystyle \frac{\sin x-\cos x}{\sin x+\cos x}$
  1. $3-\sqrt {2}$
  2. $2+\sqrt {3}$
  3. $2-\sqrt {3}$
  4. $3+\sqrt {2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The infinite series sin^2x + sin^4x + ... is geometric and sums to tan^2x. Thus 2^(tan^2x) is a root of x^2 - 9x + 8 = 0; since it is greater than 1, it equals 8, giving tan^2x = 3 and x = pi/3. Therefore, (sin x - cos x)/(sin x + cos x) = 2 - sqrt(3).