Quantitative Aptitude

Algebra

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Algebra Questions

Multiple choice

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - (1 - 2a^2) x + (1 - 2a^2) = 0$.Under what condition is $\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} < 1.$
  1. $a^2 < \dfrac{1}{2}$
  2. $a^2 > \dfrac{1}{2}$
  3. $a^2 > 1$
  4. $a^2 \varepsilon \left( \dfrac{1}{3}, \dfrac{1}{2} \right)$ only
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Roots alpha, beta satisfy alpha+beta = 1-2a^2 and alpha*beta = 1-2a^2. The expression 1/alpha^2 + 1/beta^2 = (alpha^2 + beta^2) / (alpha*beta)^2 = ((alpha+beta)^2 - 2alpha*beta) / (alpha*beta)^2. Substituting: ((1-2a^2)^2 - 2(1-2a^2)) / (1-2a^2)^2 = 1 - 2/(1-2a^2). For this to be < 1, -2/(1-2a^2) < 0, which implies 1-2a^2 > 0, so a^2 < 1/2.

Multiple choice

Let $\sin x$ and $\sin y$ be roots of the quadratic equation $a\sin^2\theta+b\sin \theta+c=0\, (a, b, c \in R$ and $a \neq 0)$ such that $\sin x+2\sin y=1$, then the value of $(a^2+2b^2+3ab+ac)$ equals
  1. $0$
  2. $1$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given sin x and sin y are roots of a*sin^2 + b*sin + c = 0. By Vieta's, sin x + sin y = -b/a and sin x * sin y = c/a. We have sin x + 2 sin y = 1. Solving these leads to the identity a^2 + 2b^2 + 3ab + ac = 0.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 4x + 1 = 0$, then the value of $\displaystyle \frac{1}{\alpha + 2 \beta} + \frac{1}{\beta + 2\alpha}$ is _________.
  1. $\displaystyle \frac{12}{13}$
  2. $\displaystyle \frac{6}{5}$
  3. $\displaystyle \frac{8}{15}$
  4. $\displaystyle \frac{12}{17}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given roots alpha and beta of 2x^2 - 4x + 1 = 0, we have alpha + beta = 2 and alpha * beta = 0.5. The expression simplifies to (alpha + beta + 2alpha + 2beta) / ((alpha + 2beta)(beta + 2alpha)) = 3(alpha + beta) / (alpha*beta + 2alpha^2 + 2beta^2 + 4alpha*beta). Using symmetric sums, the result is 12/17.

Multiple choice

If ${ \lambda }{ 1 }$ and ${ \lambda }{ 2 }$ are the two values of $\lambda$ such that the roots $\alpha$ and $\beta$ of the quadratic equation, $\lambda \left( { x }^{ 2 }-x \right) +x+5=0$ satisfy $\cfrac { \alpha }{ \beta } +\cfrac { \beta }{ \alpha } +\cfrac { 4 }{ 5 } =0$, then $\cfrac { { \lambda }{ 1 } }{ { \lambda }{ 2 }^{ 2 } } +\cfrac { { \lambda }{ 2 } }{ { \lambda }{ 1 }^{ 2 } } =0$ is equal to:
  1. $536$
  2. $512$
  3. $504$
  4. $488$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation is lambda*x^2 - lambda*x + 1 + 5 = 0, which is lambda*x^2 - lambda*x + 6 = 0. Sum of roots alpha+beta = 1, product alpha*beta = 6/lambda. (alpha/beta + beta/alpha) = ((alpha+beta)^2 - 2*alpha*beta) / (alpha*beta) = (1 - 12/lambda) / (6/lambda) = (lambda-12)/6. Given (lambda-12)/6 + 4/5 = 0, lambda-12 = -24/5, lambda = 12 - 4.8 = 7.2. This leads to a specific value for the expression.

Multiple choice

In the Maths test two representatives, while solving a quadratic equation, committed the following mistakes: (i) One of them made a mistake in the constant term and got the roots as $5$ and $9$. (ii) Another one committed an error in the coefficient of x and got the roots as $12$ and $4$. But in the meantime, they realised that they are wrong and they managed to get right jointly. Find the correct quadratic equation.
  1. $x^2+4x+14=0$
  2. $2x^2+7x-24=0$
  3. $x^2-14x+48=0$
  4. $3x^2-17x+52=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The first person got roots 5 and 9, so the sum is 14 and product is 45; since they erred in the constant, the coefficient of x (-14) is correct. The second person got roots 12 and 4, so the product is 48 and sum is 16; since they erred in the x-coefficient, the constant term (48) is correct. The correct equation is x^2 - 14x + 48 = 0.