Embibe Experts Solutions for Chapter: Quadratic Equations, Exercise 1: Exercise-1

Let $x^2-2\left(a-1\right)x+a-1=0$ where$a\in R$ be a quadratic equation, then find the number of value(s) of $a$ for which both the roots are negative. 

Let $x^2-2\left(a-1\right)x+a-1=0$ where $a\in \mathrm{R}$ be a quadratic equation, then the value(s) of $a$ for which both the roots are opposite in sign lies in the interval $\left(-\infty ,\beta \right)$. Find the value of $\beta .$

If $x^2-2\left(a-1\right)x+a-1=0$, where $a\in R$ be a quadratic equation, then find the number of value(s) of $a$ for which both the roots are greater than $1$.

If $x^2-2(a-1)x+a-1=0$, where $a\in R$, be a quadratic equation, then the set of values of $a$ for which both the roots are smaller than $1$ is$(-\infty ,\beta )$. Find $\beta .$

If $x^2-2\left(a-1\right)x+a-1=0$, where $a\in R$ be a quadratic equation, then the set of values of $a$ for which one root is smaller than $1$ and the other root is greater than $1$ is $\left(\beta ,\infty \right)$. Find $\beta$.

If the sum of the roots of the quadratic equation $\left(a+1\right)x^2+\left(2a+3\right)x+\left(3a+4\right)=0$ is $-1,$ then find the product of the roots.

Find the least prime integral value of $2a$ such that the roots $\alpha ,\beta$ of the equation $2x^2+6x+a=0$ satisfy the inequality $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }<2$.

If $\alpha , \beta , \gamma$ are the roots of the equation $x^3+p{x}^2+qx+r=0$ and the value of $\left(\alpha -\frac{1}{\beta \gamma }\right)\left(\beta -\frac{1}{\gamma \alpha }\right)\left(\gamma -\frac{1}{\alpha \beta }\right)$ is $-\frac{{\left(r-a\right)}^b}{r^c}$. Find $a+b+c$.