Question

Find the sum of the roots if \[k > 0\] and the product of the roots of the equation \[{x^2} - 3kx + 2{e^{2\log k}} - 1 = 0\] is \[7\].

Answer 1

Hint: The given equation is a quadratic equation. The general quadratic equation is \[a{x^2} + bx + c = 0\]. The sum of the roots of a quadratic equation is given by \[ - \dfrac{b}{a}\]. The product of the roots of a quadratic equation is given by \[\dfrac{c}{a}\]. Identifying the values of a, b and c from the given quadratic equation. Then substituting the values in the given general forms and simplify. We get the value of k. Now again substituting all the values in the given general form of sum of the roots of a quadratic equation.

Complete step-by-step solution:
Since the product of the given roots is \[7\].
We know that the product of the roots of a quadratic equation is given by \[\dfrac{c}{a} = 7\],
Here product of the roots is \[2{e^{2\log k}} - 1 = 7\]
\[ \Rightarrow 2{e^{2\log k}} = 8\]
\[ \Rightarrow {e^{2\log k}} = 4\]
Taking natural log on both sides,
$\Rightarrow$\[2\log k = \log 4\]
$\Rightarrow$\[2\log k = \log {2^2}\]
We know that \[\log {a^b} = b\log a\]
$\Rightarrow$\[2\log k = 2\log 2\]
Cancelling out the common terms,
$\Rightarrow$\[\log k = \log 2\]
Cancelling out the log on both sides,
$\Rightarrow$\[k = 2\]
Therefore, the sum of the roots are given by \[ - \dfrac{b}{a}\]
Here \[b = - 3k\] and \[a = 1\]
\[ \Rightarrow - \dfrac{{\left( { - 3k} \right)}}{1} = 3k = 3\left( 2 \right) = 6\]

The sum of the roots is option C.

Note: A root of a polynomial is where the polynomial is equal to zero. More specifically, a root is the x-value where the y-values are equal to zero. The general quadratic equation is \[a{x^2} + bx + c = 0\], where a, b and c are constants with \[a \ne 0\]. The roots of the quadratic equation is given by a formula:
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[x = \dfrac{{ - b}}{{2a}} \pm \dfrac{{ \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Where \[ \pm \sqrt {{b^2} - 4ac} \]is the discriminant of the equation \[a{x^2} + bx + c = 0\]
If the roots of this equation are given by \[\alpha \] and \[\beta \], then the relationships are given by,
\[\alpha + \beta = - \dfrac{b}{a}\]
Sum of the roots are given by the negative ratio of coefficient of \[x\]and \[{x^2}\]
\[\alpha \beta = \dfrac{c}{a}\]
Product of the roots are given by the ratio of the constant term and the coefficient of \[{x^2}\]