Question

If $b < 0$then the roots ${x_1}{\text{ and }}{x_2}$ of the equation $2{x^2} + 6x + b = 0$ satisfy the condition that $\left( {\dfrac{{{x_1}}}{{{x_2}}}} \right) + \left( {\dfrac{{{x_2}}}{{{x_1}}}} \right) < K$ where $K$ is equal to:
A. $2$
B. $ - 2$
C. $0$
D. $4$

Answer 1

Hint:
If we are given the quadratic equation $a{x^2} + bx + c = 0$ then we can say that for this equation we have
Sum of roots$ = - \dfrac{b}{a}$
Product of roots$ = \dfrac{c}{a}$
Using this formula we can get our answer.

Complete step by step solution:
Now we are given the quadratic equation $2{x^2} + 6x + b = 0$ and also it has two roots ${x_1}{\text{ and }}{x_2}$ and also we know that$b < 0$. So as we know that if we are given the quadratic equation $a{x^2} + bx + c = 0$ the roots are two and they can be equal or unequal depending on whether they have the perfect square or not then we can say that for this equation we have
Sum of roots$ = - \dfrac{b}{a}$$ = = - \dfrac{{{\text{coffecient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$
Product of roots$ = \dfrac{c}{a}$$ = \dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$
So for the equation $2{x^2} + 6x + b = 0$ we can say that
${x_1} + {x_2} = \dfrac{{ - 6}}{2}$
${x_1} + {x_2} = - 3$
${x_1}.{x_2} = \dfrac{b}{2}$
And we know for the real roots $D > 0$
$
  {b^2} - 4ac > 0 \\
  {6^2} - 4(2)(b) > 0 \\
  36 - 8b > 0 \\
 $
Now we are given that the roots satisfy the equation:
$\left( {\dfrac{{{x_1}}}{{{x_2}}}} \right) + \left( {\dfrac{{{x_2}}}{{{x_1}}}} \right) < K$
Taking the LCM we get that
$\left( {\dfrac{{{x_1}^2 + {x_2}^2}}{{{x_1}{x_2}}}} \right) < K - - - - - (1)$
We know that
${a^2} + {b^2} = {(a + b)^2} - 2ab$
So we can write (1) as
${x_1}^2 + {x_2}^2 = {({x_1} + {x_2})^2} - 2{x_1}{x_2}$
Now we know that ${x_1} + {x_2} = - 3$
${x_1}.{x_2} = \dfrac{b}{2}$
So putting their values we get that
$
  \left( {\dfrac{{{{({x_1} + {x_2})}^2} - 2{x_1}{x_2}}}{{{x_1}{x_2}}}} \right) < K \\
\Rightarrow \left( {\dfrac{{{{( - 3)}^2} - 2\dfrac{b}{2}}}{{\dfrac{b}{2}}}} \right) < K \\
\Rightarrow \dfrac{{18}}{b} - 2 < K \\
 $
And we are given that $b < 0$ so we can say that $\dfrac{{18}}{b} - 2$ is always less than $ - 2$ and as $b < 0$ so $\dfrac{{18}}{b}$ must also be negative so we can say that
$\dfrac{{18}}{b} - 2 < - 2$
$\dfrac{{18}}{b} - 2 < K$
On comparing the above two we get that

$K = - 2$

Note:
For any quadratic equation $f(x) = $$a{x^2} + bx + c = 0$ if $f(x)$ is positive then it must follow the certain condition that is $a > 0$ and $D < 0$ where $D = {b^2} - 4ac$ so we can say for $a > 0$$D = {b^2} - 4ac < 0$ $f(x)$ will always be positive.