We can solve this using the very common ‘find the determinant and equate to zero method’
From there, I got the equation $$2k^2-30k=0$$ $$k=0,15$$ $k=0$ Won’t apply because then it wouldn’t be a pair of straight line. Why did we get an extraneous answer?
Unfortunately, the use of the discrminant $\Delta =0$ for determining whether pair of lines exist is not very good. See both $x^2+1=0$ and $x^2-1=0$ give $\Delta =0$ but only latter has (Real) pair of straight lines. Therefore one should also use the algebraic ideas likereal-factoriztion etc. This is what @Piquito's solution also d0es.
Yes $k=15$ will do: $15xy+10x+6y+4=0 \implies (5x+2)(3y+2)=0 \implies x=-2/5, y=-2/3 $$ (two lines)
Your answer is not extraneous. Simply with the value $k=0$ you do have only an unique straight line (which with your method of solution must be interpreted as two overlapping straight lines). Another way of finding $k=15$ could be as follows:
Since there are not coefficients of $x^2$ nor $y^2$ the correspondig product is of the form $$(ax+b)(cy+d)=acxy+adx+bcy+bd$$ then you can eliminate $a$ and $b$ so you have $$a=\frac kc=\frac {10}{d}\\b=\frac 6c=\frac 4d$$ from which $$kd=10c\\6d=4c$$ Thus $k=15$.
Note that the determinate is defined for conic (quadratic) equations, which excludes the case of $k=0$. So, you should assume $k\ne 0 $ if you compute the determinate of the given function.
Maybe a cleaner way to find $k$ is to rearrange the equation as
$$kxy+10x+6y+4=k\left(xy+\frac{10}kx+\frac6ky\right)+4 =k\left(x+\frac{6}k\right)\left(y+\frac{10}k\right)+4-\frac{60}k=0$$
or,
$$k\left(x+\frac{6}k\right)\left(y+\frac{10}k\right)=\frac{60}k-4$$
Then, set the RHS to zero to obtain $k=15$.
The second solution isn’t extraneous. Finding values of $k$ where the determinant vanishes gets you degenerate conics. The problem is that there are other types of degenerate conics besides a pair of distinct intersecting lines. You might have a pair of parallel lines, a single line, a single point, or no real solution at all. To select among these possibilities you also have to examine at least the equation’s discriminant.
In this case, the discriminant is equal to $-(k/2)^2$. For a pair of distinct intersecting lines, we want this to be negative, so that the conic is a degenerate hyperbola. This corresponds to the solution $k=15$ of your determinant equation. When $k=0$, the discriminant is zero, which means that you have a degenerate parabola, here the single line $10x+6y+4=0$.
