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 $\alpha$ and $ \beta$ are the roots of the equation $ { x^{ 2 } }-ax+b=0$ and $ { v }_{ n }={ \alpha }^{ n }+{ \beta }^{ n }$, then
  1. $ { v }_{ n+1 }={ av }_{ n }-{ bv }_{ n }$
  2. $ { v }_{ n+1 }={ bv }_{ n }-{ av }_{ n-1 }$
  3. $ { v }_{ n+1 }={ av }_{ n }-{ bv }_{ n-1 }$

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

Given the quadratic equation x^2 - ax + b = 0, the roots alpha and beta satisfy alpha + beta = a and alpha * beta = b. Since alpha and beta are roots, they satisfy alpha^2 = a*alpha - b and beta^2 = a*beta - b. Multiplying by alpha^(n-1) and beta^(n-1) respectively and adding gives v_{n+1} = a*v_n - b*v_{n-1}.

Multiple choice

The value of $p$ for which one root of the quadratic equation $\displaystyle \left ( p^{2}-5p+3 \right )x^{2}+\left ( 3p-1 \right )x+2= 0$ is twice as large as the other is
  1. $\displaystyle -\frac{2}{3}$
  2. $\displaystyle \frac{1}{3}$
  3. $\displaystyle -\frac{1}{3}$
  4. $\displaystyle \frac{2}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let roots be r and 2r. Sum = 3r = -(3p-1)/(p^2-5p+3). Product = 2r^2 = 2/(p^2-5p+3). From product, r^2 = 1/(p^2-5p+3). Square the sum: 9r^2 = (3p-1)^2 / (p^2-5p+3)^2. Substitute r^2: 9 / (p^2-5p+3) = (3p-1)^2 / (p^2-5p+3)^2. 9 = (3p-1)^2 / (p^2-5p+3). 9(p^2-5p+3) = 9p^2-6p+1. 9p^2-45p+27 = 9p^2-6p+1. 39p = 26. p = 26/39 = 2/3.

Multiple choice

Let $a, b, c$ be the lengths of sides of a scalene triangle. If the roots of the equation $\displaystyle x^{2}+2(a+b+c)x+3\lambda (ab+bc+ca)=0$ $(\lambda \in R)$ are real then
  1. $\displaystyle \lambda <\dfrac{4}{3}$
  2. $\displaystyle \lambda >\frac{5}{3}$
  3. $\displaystyle \lambda \in \left ( \frac{1}{3}, \frac{5}{3} \right )$
  4. $\displaystyle \lambda \in \left ( \frac{4}{3}, \frac{5}{3} \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For roots to be real, discriminant D >= 0. D = [2(a+b+c)]^2 - 4(1)(3*lambda*(ab+bc+ca)) >= 0. 4(a+b+c)^2 - 12*lambda*(ab+bc+ca) >= 0. (a^2+b^2+c^2+2ab+2bc+2ca) - 3*lambda*(ab+bc+ca) >= 0. Since a, b, c are sides of a triangle, we use the property that (a+b+c)^2 > 3(ab+bc+ca). The condition leads to lambda < 4/3.

Multiple choice

If the roots of equation $ a\left( b-c \right) { x }^{ 2 }+b\left( c-a \right) x+c\left( a-b \right) =0$ be equal, then
  1. $2b= a + c$
  2. $b^2 = ac$
  3. $\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$

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

For equal roots, the discriminant b^2 - 4ac = 0. Here, A = a(b-c), B = b(c-a), C = c(a-b). B^2 - 4AC = 0 leads to [b(c-a)]^2 - 4[a(b-c)][c(a-b)] = 0. This simplifies to b^2(c-a)^2 + 4ac(b-c)(a-b) = 0. Expanding and rearranging leads to the condition 2/b = 1/a + 1/c.

Multiple choice

If the roots of the equation $\displaystyle ax^{2} + bx + c = 0$ are equal and $\displaystyle f\left ( x \right ) = \begin{vmatrix}ax+b & x^{2} & 4a \ bx+c & x & b \ cx+a & 1 & 4c\end{vmatrix}$, then the possible roots of $\displaystyle f\left ( x \right ) = 0$ is/are
  1. $\displaystyle 0$
  2. $\displaystyle 1$
  3. $\displaystyle -1$

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

If roots of ax^2 + bx + c = 0 are equal, then b^2 = 4ac. The determinant f(x) simplifies. Given the structure, x=0 is a common root for such determinant equations.

Multiple choice

If rational numbers $a,b,c,d$ are in G.P., then roots of equation $\displaystyle (a-c)^{2}x^{2}+(b-c)^{2}x+(b-d)^{2}= (a-d)^{2} $ are necessarily

  1. Imaginary

  2. Irrational

  3. rational

  4. real and distinct
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If a, b, c, d are in G.P., let b=ar, c=ar^2, d=ar^3. Substituting these into the equation (a-c)^2 x^2 + (b-c)^2 x + (b-d)^2 = (a-d)^2 and simplifying shows the coefficients are rational expressions of a and r. Since a, b, c, d are rational, the coefficients are rational, leading to rational roots.

Multiple choice

The value of $P$ for which both the roots of the equation $\displaystyle 4x^{2}-20Px+(25P^{2}+15P-66)= 0$ are less than $2$, lies in
  1. $\displaystyle \left ( \frac{4}{5},2 \right )$
  2. $\displaystyle (2,\infty )$
  3. $\displaystyle \left (-1, -\frac{4}{5} \right )$
  4. $\displaystyle ( -\infty,-1)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For roots to be less than 2, the vertex -b/2a must be < 2, f(2) > 0, and D >= 0. Solving these inequalities for P leads to P < -1.

Multiple choice

The roots of the quadratic equation $\sqrt{3}x^{2}+10x+7\sqrt{3}=0$ are
  1. $\left [ -\sqrt{3},-\frac{7}{\sqrt{3}} \right ]$
  2. $\left [ \frac{1}{\sqrt{3}},\frac{\sqrt{3}}{7} \right ]$
  3. $\left [ \sqrt{3},\frac{7}{\sqrt{3}} \right ]$
  4. $\left [ \sqrt{3},-\frac{7}{\sqrt{3}} \right ]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The quadratic equation can be factored by splitting the middle term: sqrt(3)x^2 + 3x + 7x + 7*sqrt(3) = 0, which simplifies to (x + sqrt(3))(sqrt(3)*x + 7) = 0. Solving for x gives the roots as -sqrt(3) and -7/sqrt(3).

Multiple choice

The value of $a$ for which one root of the quadratic equation $\displaystyle (a^{2}-5a+3)x^{2}+(3a-1)x+2= 0$ is twice as large as the other is
  1. $\displaystyle \frac{2}{3}$
  2. $\displaystyle \frac{1}{3}$
  3. $\displaystyle \frac{-2}{3}$
  4. $\displaystyle \frac{-1}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let roots be r and 2r. Sum = 3r = -(3a-1)/(a^2-5a+3). Product = 2r^2 = 2/(a^2-5a+3). From product, r^2 = 1/(a^2-5a+3). Squaring the sum: 9r^2 = (3a-1)^2 / (a^2-5a+3)^2. 9/(a^2-5a+3) = (3a-1)^2 / (a^2-5a+3)^2. 9 = (3a-1)^2 / (a^2-5a+3). 9(a^2-5a+3) = 9a^2 - 6a + 1. 9a^2 - 45a + 27 = 9a^2 - 6a + 1. 39a = 26. a = 26/39 = 2/3.

Multiple choice

If $\displaystyle x^{2}+(a-b)x+(1-a-b)= 0, $ where $\displaystyle a,b \in R,$ the value of $a$ such that the equation has distinct real roots for all value of $b$ are
  1. $\displaystyle a> 1 $
  2. $\displaystyle a< 1 $
  3. $\displaystyle a> 2 $
  4. $\displaystyle a< 2 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For distinct real roots, the discriminant D = (a-b)^2 - 4(1-a-b) > 0. Expanding: a^2 - 2ab + b^2 - 4 + 4a + 4b > 0. This must hold for all b, which implies the quadratic in b must have a negative discriminant.

Multiple choice

If the roots of the equation $\displaystyle bx^{2}+cx+a= 0$ be imaginary, then for all real values of $x$. The expression $\displaystyle 3b^{2}x^{2}+6bcx+2c^{2}$ is
  1. less than $4ab$
  2. greater than $-4ab$
  3. less than $-4ab$
  4. greater than $4ab$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If roots of bx^2 + cx + a = 0 are imaginary, the discriminant c^2 - 4ab < 0, so c^2 < 4ab. The expression 3b^2x^2 + 6bcx + 2c^2 can be written as 3(bx + c)^2 - c^2. Since c^2 < 4ab, -c^2 > -4ab. This logic is complex, but testing values confirms the inequality.

Multiple choice

If $ a+b+c > 0 $, $a < 0 < b < c$, then roots of the equation $\displaystyle a\left ( x-b \right )\left ( x-c \right )+b \left ( x-c \right )\left ( x-a \right )+c\left ( x-a \right )\left ( x-b \right )=0 $ are real.

  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
A Correct answer
Multiple choice

The quadratic equations $\displaystyle x^{2}-6x+a= 0$ and $\displaystyle x^{2}-cx+6= 0$ have one root in common. The other roots of the first & the second equations are integers in the ratio $4 : 3$. then the common root is
  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let common root be k. k^2 - 6k + a = 0 and k^2 - ck + 6 = 0. Other roots are r1 = 6/k and r2 = a/k. Ratio r1/r2 = 4/3 or 3/4. If k=2, r1=3, r2=a/2. If r1/r2 = 3/4, 3/(a/2) = 3/4 => a/2=4 => a=8. Check: x^2-6x+8=(x-2)(x-4), x^2-cx+6=(x-2)(x-3). Common root is 2.

Multiple choice

The number of points $(p, q)$ such that $p, q\in \left { 1, 2, 3, 4 \right }$ and the equation $px^{2}+qx+1=0$ has real roots is
  1. $7$
  2. $8$
  3. $9$

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

For real roots, the discriminant D = q^2 - 4p >= 0, so q^2 >= 4p. Testing values for p, q in {1, 2, 3, 4}: If p=1, q^2 >= 4 (q=2,3,4 -> 3 pairs). If p=2, q^2 >= 8 (q=3,4 -> 2 pairs). If p=3, q^2 >= 12 (q=4 -> 1 pair). If p=4, q^2 >= 16 (q=4 -> 1 pair). Total = 3 + 2 + 1 + 1 = 7.