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 arrangement of the following quadratic equations in the descending order of their sum of the roots. A: $ x^{2}+11=0$ B: $(x+2) (x-3) =0$ C: $5x^{ 2 }+4x=0$ D: $2x^{ 2 }+3x-5=0$
  1. $A,B,C,D$
  2. $D,C,B,A$
  3. $B,A,C,D$
  4. $A,C,B,D$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Sum of roots for ax^2 + bx + c = 0 is -b/a. A: x^2 + 11 = 0, sum = 0. B: x^2 - x - 6 = 0, sum = 1. C: 5x^2 + 4x = 0, sum = -4/5 = -0.8. D: 2x^2 + 3x - 5 = 0, sum = -3/2 = -1.5. Descending order: B(1) > A(0) > C(-0.8) > D(-1.5).

Multiple choice

A quadratic equation with integral coefficient has two different prime numbers as its roots if the sum of the coefficients of the equation is prime, then the sum of the roots is
  1. $2$
  2. $5$
  3. $7$
  4. $11$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the roots be p and q (primes). The quadratic is (x-p)(x-q) = x^2 - (p+q)x + pq = 0. The sum of coefficients is 1 - (p+q) + pq = prime. This factors as (p-1)(q-1) = prime. Since p and q are primes, the only way for (p-1)(q-1) to be prime is if one factor is 1. Thus, p-1 = 1, so p = 2. Then (2-1)(q-1) = q-1 = prime. If q=3, q-1=2 (prime). Roots are 2 and 3. Sum = 5.

Multiple choice

If one root of the quadratic equation $2x^{2}-2kx+k-4=0$ is smaller than $1$ & other is greater than $2$, then the complete set of values of $k$ is
  1. $\left ( -2,\infty \right )$
  2. $\left ( \dfrac{4}{3},\infty \right )$
  3. $\left (-2,\dfrac{4}{3} \right )$
  4. $\left ( -\infty ,-2 \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let f(x) = 2x^2 - 2kx + k - 4. For one root < 1 and other > 2, the conditions are f(1) < 0 and f(2) < 0. f(1) = 2 - 2k + k - 4 = -k - 2 < 0 => k > -2. f(2) = 8 - 4k + k - 4 = 4 - 3k < 0 => k > 4/3. Intersection is k > 4/3.

Multiple choice

Let two circles with radii $r_1$ and $r_2$ have their centers at a distance of $ \sqrt{2} \left(r_1-r_2 \right)$. If $r_1$ and $r_2$ are roots of the equation $x^2 - 2(\alpha + \beta)x + \alpha^2 + \beta^2=0$, then the relation between $\alpha$ and $\beta$ for which the two circles are orthogonal is
  1. $\alpha^{ 2} + \beta^{2} =4\alpha \beta $
  2. $(\alpha +\beta)^{2}=4\alpha \beta $
  3. $\alpha ^{2}+\beta ^{2}=\alpha \beta $
  4. $\alpha ^{2}-\beta ^{2}=4\alpha \beta$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For orthogonal circles, 2*g1*g2 + 2*f1*f2 = c1 + c2. Given the distance between centers is sqrt(2)(r1-r2), the condition simplifies to the relation between radii and centers. The roots of x^2 - 2(a+b)x + a^2+b^2 = 0 are r1, r2. The condition for orthogonality leads to a^2 + b^2 = 4ab.

Multiple choice

If $\displaystyle \alpha, : \beta, : \gamma$ are roots of the equation $\displaystyle x^{3} - ax^{2} + bx - c = 0$, then the value of the expression $\displaystyle \left ( \alpha + \beta - \gamma \right ) \left ( \beta + \gamma - \alpha \right )\left ( \gamma + \alpha - \beta \right )$ equals
  1. $\displaystyle 4ab - 8c$
  2. $\displaystyle a^{3} - 8c$
  3. $\displaystyle 4ab - a^{3} - 8c$
  4. $\displaystyle a^{3} -8c - 4ab$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Vieta's formulas: alpha+beta+gamma = a, alphabeta+betagamma+gammaalpha = b, alphabetagamma = c. The expression expands to (a-2gamma)(a-2alpha)(a-2beta). This evaluates to a^3 - 2a^2(alpha+beta+gamma) + 4a(alphabeta+betagamma+gammaalpha) - 8alphabetagamma = a^3 - 2a^3 + 4ab - 8c = 4ab - a^3 - 8c.

Multiple choice

If $\vec {a}=2\hat {i}-\hat {j}+\hat {k},:\vec {b}=\hat {i}+2\hat {j}-3\hat {k},:\ \vec {c}=3\hat {i}+\mu\hat {j}+5\hat {k}$ are coplanar, then $\mu$ is a root of the equation
  1. $x^{2}+3x=4$
  2. $x^{2}+2x=6$
  3. $x^{2}+3x=6$

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

Three vectors are coplanar if their scalar triple product is zero. The determinant of the matrix formed by the components is 0. Solving the determinant yields a quadratic equation in mu, which matches the form x^2 + 3x = 4.

Multiple choice

If $\alpha ,{ \alpha }^{ 2 }$ are the roots of the equation ${ x }^{ 2 }-6x+c=0$, then the positive value of $c$ is
  1. $2$
  2. $3$
  3. $4$
  4. $9$
  5. $8$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For the equation x^2 - 6x + c = 0, the sum of roots is alpha + alpha^2 = 6 and the product is alpha * alpha^2 = c. Solving alpha^2 + alpha - 6 = 0 gives (alpha + 3)(alpha - 2) = 0, so alpha is 2 or -3. If alpha = 2, c = 2^3 = 8; if alpha = -3, c = (-3)^3 = -27. Since c must be positive, c = 8.

Multiple choice

The condition that a roots of the equation $ax^{2}+bx+c=0$ may be reciprocal to the roots of $a_{1}x^{2}+b_{1}x +c_{1}=0$ is
  1. $(bb_{1} -aa_{1})^{2}=(ab_{1}-bc_{1})(ba_{1}-b_{1}c)$
  2. $(cb_{1}-ba_{1})^{2}=(ac_{1}-cd_{1})(ab_{1}-bc_{1})$
  3. $(cc_{1}-aa_{1})^{2}=(ab_{1}-bc_{1})(ba_{1}-b_{1}c)$
  4. ${a}+{b}+{c}=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A root reciprocal to a root of the second equation is a common root of ax^2+bx+c and c1x^2+b1x+a1. Their resultant must be zero, which gives (cc1-aa1)^2 = (ab1-bc1)(ba1-b1c). This is the condition shown in option C.

Multiple choice

I: If $x+2$ is factor of $x^{2}+2ax+b$ and $x^{2}+2cx+d$, then $\dfrac{b-d}{a-c}=4$ II: The equation whose roots are the squares of the roots of $x^{2}+4x+5=0$ is $x^{2}-6x+25=0$ Which of the above statement(s) is/are true?

  1. Only I

  2. Only II

  3. Both I and II

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

For I: If x+2 is a factor, then f(-2)=0. So 4-4a+b=0 and 4-4c+d=0. Thus b=4a-4 and d=4c-4. (b-d)/(a-c) = (4a-4c)/(a-c) = 4. For II: If roots of x^2+4x+5=0 are alpha, beta, then alpha+beta=-4 and alpha*beta=5. The new roots are alpha^2, beta^2. Sum = (alpha+beta)^2 - 2alpha*beta = 16-10=6. Product = (alpha*beta)^2 = 25. Equation is x^2-6x+25=0. Both are true.