Question

The product of the roots of the equation \[5{x^2} - 4x + 2 + k(4{x^2} - 2x - 1) = 0\] is 2 then find the value of k
A) \[ - \dfrac{8}{9}\]
B) \[\dfrac{8}{9}\]
C) \[ - \dfrac{4}{9}\]
D) \[\dfrac{4}{9}\]

Answer 1

Hint: If we let the roots of a certain quadratic equation \[a{x^2} + bx + c = 0\] to be \[\alpha \& \beta \] then it must be remembered that \[\alpha + \beta = - \dfrac{b}{a}\] and \[\alpha \beta = \dfrac{c}{a}\] . Using this we can solve this equation.

Complete step-by-step answer:
First of all let us try to simplify this equation in such a way that we can compare it with the general quadratic equation i.e., \[a{x^2} + bx + c = 0\] For this let us try to solve the given equation further
We will get it as,
\[\begin{array}{l}
\therefore 5{x^2} - 4x + 2 + k(4{x^2} - 2x - 1) = 0\\
 \Rightarrow 5{x^2} - 4x + 2 + 4k{x^2} - 2kx - k = 0\\
 \Rightarrow {x^2}(5 + 4k) - x(4 + 2k) + (2 - k) = 0
\end{array}\]
Now here we can compare the coefficients of \[{x^2}\] ,\[x\] and the constant part also
\[\begin{array}{l}
a = (5 + 4k)\\
b = - (4 + 2k)\\
c = (2 - k)
\end{array}\]
 Now as we have the values of a, b, c . We know that \[\alpha \beta = \dfrac{c}{a}\]
Where \[\alpha \& \beta \] are the roots of the equation.
And as it is also given in the question that the product of roots of the equation is
\[\alpha \beta = \dfrac{c}{a} = 2\]
Now putting the value of c and a we get
\[\begin{array}{l}
\therefore \alpha \beta = \dfrac{c}{a} = \dfrac{{2 - k}}{{5 + 4k}} = 2\\
 \Rightarrow 2 - k = 2(5 + 4k)\\
 \Rightarrow 2 - k = 10 + 8k\\
 \Rightarrow 2 - 10 = 8k + k\\
 \Rightarrow 9k = - 8\\
 \Rightarrow k = - \dfrac{8}{9}
\end{array}\]
So clearly option A is the correct option.

Note: Always use what's given in the question, the product of the roots were already given; don't try to find the roots by sridharacharya formula, just apply the formula to solve it in the last possible time.