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

Find a quadratic polynomial whose zeros are $5 + \sqrt 2$ and $5- \sqrt 2$.
  1. $k \{x^2 - 2 x + 23 \}$, where $k$ is any non-zero real number.
  2. $k \{x^2 - 16 x + 23 \}$, where $k$ is any non-zero real number.
  3. $k \{x^2 - 10 x + 23 \}$, where $k$ is any non-zero real number.
  4. $k \{x^2 - 14 x + 23 \}$, where $k$ is any non-zero real number.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Sum of zeros = (5 + sqrt(2)) + (5 - sqrt(2)) = 10. Product of zeros = (5 + sqrt(2))(5 - sqrt(2)) = 25 - 2 = 23. Polynomial is k(x^2 - (sum)x + product) = k(x^2 - 10x + 23).

Multiple choice

A quadratic equation, whose zeroes are $-3$ and $4$, is
  1. $x^2-x+12=0$
  2. $x^2+x+12=0$
  3. $\cfrac{x^2}{2}-\cfrac{x}{2}-6=0$
  4. $2x^2+2x-24=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A quadratic equation with roots r1 and r2 is x^2 - (r1+r2)x + (r1*r2) = 0. Here, sum = -3+4 = 1, product = -3*4 = -12. Equation: x^2 - x - 12 = 0. Option C is (1/2)x^2 - (1/2)x - 6 = 0, which is equivalent to (1/2)(x^2 - x - 12) = 0.

Multiple choice

$\alpha$ and $\beta$ are two numbers such that $\alpha + \beta = 6$, $\alpha - \beta = 8$. Then, $\alpha$ and $\beta$ are the roots of the quadratic equation __________.
  1. $x^2 + 6x + 7 = 0$
  2. $x^2 - 6x - 7 = 0$
  3. $x^2 + 6x - 8 = 0$
  4. $x^2- 6x + 8 = 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A quadratic equation with roots alpha and beta is given by x^2 - (alpha + beta)x + alpha * beta = 0. We are given the sum of the roots alpha + beta = 6. Solving the system alpha + beta = 6 and alpha - beta = 8 gives alpha = 7 and beta = -1, which means the product of the roots is -7. Substituting these values into the formula yields x^2 - 6x - 7 = 0.

Multiple choice

Sum and product of the zeroes a quadratic polynomial are given. Find the zeroes of the polynomial by factorisation. $\dfrac{-3}{2\sqrt{5}},-\dfrac{1}{2}$
  1. $\dfrac{\sqrt{5}}{3},\dfrac{-\sqrt{5}}{2}$
  2. $\dfrac{\sqrt{5}}{5},\dfrac{-\sqrt{5}}{2}$
  3. $\dfrac{\sqrt{5}}{7},\dfrac{-\sqrt{5}}{2}$
  4. $\dfrac{\sqrt{5}}{6},\dfrac{-\sqrt{5}}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Sum = -3/(2*sqrt(5)), Product = -1/2. The quadratic is x^2 - (sum)x + (product) = 0. x^2 + (3/(2*sqrt(5)))x - 1/2 = 0. Roots are given by quadratic formula. Testing option B: sum = sqrt(5)/5 - sqrt(5)/2 = (2*sqrt(5) - 5*sqrt(5))/10 = -3*sqrt(5)/10 = -3/(2*sqrt(5)). Product = (sqrt(5)/5) * (-sqrt(5)/2) = -5/10 = -1/2.

Multiple choice

If $\alpha, \beta$ are the zeroes of the quadratic polynomial $f(x)=x^2-5x+4$, then $\dfrac {1}{\alpha}+\dfrac {1}{\beta}-2\alpha \beta=$
  1. $\dfrac {27}{4}$
  2. $-\dfrac {27}{4}$
  3. $\dfrac {4}{27}$
  4. $-\dfrac {4}{27}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For x^2 - 5x + 4 = 0, the sum of roots alpha + beta = 5 and product alpha * beta = 4. The expression 1/alpha + 1/beta - 2*alpha*beta simplifies to (alpha + beta)/(alpha * beta) - 2*alpha*beta, which is 5/4 - 2(4) = 5/4 - 8 = (5 - 32)/4 = -27/4.

Multiple choice

If $2, 3$ are the zeroes of a quadratic polynomial, then polynomial is $x^2-5x+6.$

  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. Assertion is incorrect but reason is correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If 2 and 3 are zeroes, the polynomial is (x-2)(x-3) = x^2 - 5x + 6. Both the assertion and the reason are correct, and the reason explains the assertion.

Multiple choice

If roots of cubic equation are in G.P. , $ ax^3 + bx^2 + cx + d $ then :
  1. ${c^3}a = {b^3}d$
  2. $a^2 c = b^2 d $
  3. $ ac^2 = bd^2 $

  4. NOTA
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots are in GP, let them be k/r, k, kr. The product of roots is -d/a = k^3. The sum of roots taken two at a time is c/a = k^2/r + k^2 + k^2r = k^2(1/r + 1 + r). The sum of roots is -b/a = k(1/r + 1 + r). Thus, c/a = (k)(-b/a), so c/a = -b/a * k. Since k = (-d/a)^(1/3), this leads to the relation c^3 * a = b^3 * d.

Multiple choice

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - (\sqrt{5} -1)x -(\sqrt{5} + 1)$, then the value of $\displaystyle \frac{1}{\alpha^2} + \frac{1}{\beta^2}$ is ____________.
  1. $3 + \sqrt{5}$
  2. $3 - \sqrt{5}$
  3. $\sqrt{5} - 3$
  4. $-3 - \sqrt{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Sum of roots alpha + beta = sqrt(5) - 1. Product alpha * beta = -(sqrt(5) + 1). 1/alpha^2 + 1/beta^2 = (alpha^2 + beta^2) / (alpha * beta)^2 = ((alpha + beta)^2 - 2 * alpha * beta) / (alpha * beta)^2. Plugging in values: ((sqrt(5)-1)^2 + 2(sqrt(5)+1)) / (-(sqrt(5)+1))^2 = (5 + 1 - 2*sqrt(5) + 2*sqrt(5) + 2) / (5 + 1 + 2*sqrt(5)) = 8 / (6 + 2*sqrt(5)) = 4 / (3 + sqrt(5)). Rationalizing gives 4(3 - sqrt(5)) / (9 - 5) = 3 - sqrt(5).