Question

If product of roots of the equation ${x^2} - 3kx + 2{e^ {\log k}} - 1 = 0$is $7$, then
A.roots are integers and positive
B.roots are integers and negative
C.roots are rational and integers
D.roots are irrational

Answer 1

Hint:- The product of the roots of a quadratic equation is equal to the constant term (the third term, divided by the leading coefficient)


Complete step by step by solution
These are called the roots of the quadratic equation.
For a quadratic equation$a{x^2} + bx + c = 0$,
The sum of this roots = $ - \dfrac{b}{a}$
And the product of this roots = $\dfrac{c} {a} $
A quadratic equation may be expressed as a product of two binomials.
Formula
$a{x^2} + bx + c = 0$
Sum of the roots = $ - \dfrac{b}{a}$
Product of the roots = $\dfrac{c} {a} $
Where a is the coefficient of ${x^2} $
b is the coefficient of x and c is the coefficient of ${x^0}$
Step – I

${x^2} - 3kx + 2{e^ {\log k}} - 1 = 0$
Compare with
$a{x^2} + bx + c = 0$
Then, we have
$a = 1, b = - 3k, c = 2{e^ {\log k}} - 1$
Products of roots given $\dfrac{c} {a} = 7$
$\dfrac{c} {a} = \dfrac{{2{e^{\log k}} - 1}} {1}................... (2)$
                    [From question]
$\dfrac{c} {a} = 7............. (3)$
Put equation $(2) $= equation $(3) $
$\dfrac{c} {a} = 7 = \dfrac{{2{e^{\log k}} - 1}} {1}............. (4)$
Solve equation $(4) $
$\begin {gathered}
  7 = 2{e^ {\log k}} - 1 \\
  7 + 1 = 2{e^ {\log k}} \\
  8 = 2{e^ {\log k}} \\
  \dfrac{8}{2} = {e^ {{{\log} _e} k}} \\
  4 = {e^ {{{\log} _e} k}} \\
  4 = k {e^ {\log e}} \\
  4 = k \\
\end{gathered} $
We know, ${e^ {{{\log} _e} a}} = a\log e = a$
D = $144 - 4(7) = 116$is not a perfect square.
$ \Rightarrow $ Roots are irrational numbers.


Note –The sum of the roots of a quadratic equation –
$\begin {gathered}
  {r_1} + {r_2} = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} + \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}} \\
   = \dfrac{{ - b + \sqrt {{b^2} - 4ac} - b - \sqrt {{b^2} - 4ac} }}{{2a}} \\
   = - \dfrac{{2b}}{{2a}} = - \dfrac{b}{a} \\
\end{gathered} $
The sum of roots of a quadratic equation = $ - \dfrac{b}{a}$
The product

\[{r_1}.{r_2} = \dfrac{{ - b + \sqrt d }}{{2a}} + \dfrac{{ - b - \sqrt d }}{{2a}}\]
\[ = \dfrac{{d - {b^2}}}{{4a}}\]
Where
$d = {b^2} - 4ac$
\[\begin{gathered}
   = \dfrac{{{b^2} - 4ac - {b^2}}}{{4{a^2}}} \\
   = \dfrac{{4ac}}{{4{a^2}}} \\
   = \dfrac{c}{a} \\
\end{gathered} \]
The product of the roots of a quadratic equation \[= \dfrac{c}{a}\]