Question

The integral values of $k$ for which the equation $(k - 2){x^2} + 8x + k + 4 = 0$ has both the roots real, distinct and negative is:

Answer 1

Hint: In the equation containing unknowns always go with the next conditions given. Real, distinct roots mean both the roots are different.

Complete step by step solution:
Given that: $(k - 2){x^2} + 8x + k + 4 = 0$
To find the value of $k$,
Given condition is such that the roots are real, distinct and negative,
Therefore, $\Delta > 0$
Comparing the given equation with the standard quadratic equations-
$a{x^2} + bx + c = 0$
$\Delta = {b^2} - 4ac$
$\begin{array}{l}
 \Rightarrow a = (k - 2)\\
 \Rightarrow b = 8\\
 \Rightarrow c = (k + 4)
\end{array}$
$\Delta > 0$
$\therefore \Delta = {b^2} - 4ac > 0$
$\begin{array}{l}
 \Rightarrow \Delta = {8^2} - 4(k - 2)(k + 4) > 0\\
 \Rightarrow \Delta = 64 - 4({k^2} + 2k - 8) > 0\\
 \Rightarrow \Delta = 64 - 4{k^2} - 8k + 32 > 0
\end{array}$
 Simplifying the above equations and finding its factors
$ \Rightarrow \Delta = - 4{k^2} - 8k + 96 > 0$
Taking $( - 4)$ common from both the sides-
$\begin{array}{l}
 \Rightarrow \Delta = {k^2} + 2k - 24 < 0\\
 \Rightarrow \Delta = {k^2} + 6k - 4k - 24 < 0\\
 \Rightarrow \Delta = (k + 6)(k - 4) < 0
\end{array}$
Therefore , $k \in ( - 6,4)$ ---(1)
Now according to the given condition both the roots are negative-
 So the product of both the roots should be positive.
In this case the graph of these points would lie in the second and third quadrant.
$ \Rightarrow \dfrac{{ - b}}{{2a}} < 0$
Put all the values,
$\begin{array}{l}
 \Rightarrow \dfrac{{ - 8}}{{2(k - 2)}} < 0\\
 \Rightarrow \dfrac{4}{{k - 2}} > 0\\
\therefore K \in (2,\infty )
\end{array}$
                                    -- (2)
Taking Intersection of the equations (1) & (2)
$\begin{array}{l}
 \Rightarrow ( - 6,4) \cap (2,\infty )\\
 \Rightarrow k \in (2,4)
\end{array}$
Thus the value of $k$ is between $2$ and $4$ , is the number $3$ .
Thus, $k = 3$ is the required answer.

Note: Every quadratic polynomial has almost two roots. Remember, the quadratic equations having coefficients as the rational numbers has the irrational roots. Also, the quadratic equations whose coefficients are all the distinct irrationals but both the roots are the rational.