How do I find the value of $k$ such that $(x-1)/(x-3) = (k-9x)/(x+2)$ has exactly one solution?

Question

The answer is supposed to be $k = 14$ or $54$ with the quadratic formula. I've tried cross multiplying and doing something like this:

$1. (x-1)(x+2) = (k-9x)(x-3)$

$2. x^2 + 2x - x - 2 = kx - 3k - 9x^2 + 27x$

$3. x^2 + 2x - x - 2 + 9x^2 - 27x = kx - 3k$

$4. 10x^2 - 26x - 2 = kx - 3k$

Then I get stuck at step $4$. Are the calculations above correct? And how would I solve for $k$ with the quadratic formula?

Answer 1

Your calculation is correct, and rewrite it as: $10x^2 - (26+k)x + 3k -2 = 0$. The exact one solution means that $\triangle = (26+k)^2 - 4(10)(3k-2) = 0$. Can you solve this equation to get $k$ ?

Answer 2

Additional to $k=14$ or $54$, when $k=-18$, the original rational equation

$$\frac{x-1}{x-3} = \frac{k-9x}{x+2} \tag 0$$

has exactly one solution $x=14/5$. The RHS of $(0)$ is constant $-9$, except when the equation is undefined at $x=-2$ or $x=3$.

This $k=-18$ can be found by substituting $x=-2$ into your

$$(x-1)(x+2) = (k-9x)(x-3) \tag 1$$

then solve for $k$. Then with $k=-18$, the quadratic equation $(1)$ has two solutions, but the solution $x=-2$ has to be rejected for the rational equation $(0)$.

Answer 3

Cross-multiplying can be risky in working with ratio-equations: it may complicate solution because of the implicit assumption that neither denominator is equal to zero. We now tend to teach that it is preferable to create a new equation where the difference of the ratios is equal to zero, thus

$$ \frac{x \ - \ 1}{x \ - \ 3} \ - \ \frac{k \ - \ 9x}{x \ + \ 2} \ \ = \ \ \frac{(x \ - \ 1)·(x \ + \ 2) \ - \ (k \ - \ 9x)·(x \ - \ 3)}{(x \ - \ 3)·(x \ + \ 2)} \ \ = \ \ 0 $$ $$ \Rightarrow \ \ \frac{10·x^2 \ - \ (k \ + \ 26)·x \ + \ (3k \ - \ 2)}{(x \ - \ 3)·(x \ + \ 2)} \ \ = \ \ 0 \ \ . $$ The quadratic in the numerator will produce only one (double) zero when its discriminant $ \ \Delta \ = \ (k + 26)^2 - 4·10·(3k - 2) \ = \ k^2 - 68k + 756 \ = \ (k - 14)·(k - 54) \ $ is zero. [The solutions are

$$ \ \mathbf{k \ = \ 14 \ \ :} \quad \quad \frac{10·x^2 \ - \ 40·x \ + \ 40}{(x \ - \ 3)·(x \ + \ 2)} \ \ = \frac{10·(x \ - \ 2)^2}{(x \ - \ 3)·(x \ + \ 2)} \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ = \ 2 \ \ ; $$

$$ \ \mathbf{k \ = \ 54 \ \ :} \quad \quad \frac{10·x^2 \ - \ 80·x \ + \ 160}{(x \ - \ 3)·(x \ + \ 2)} \ \ = \frac{10·(x \ - \ 4)^2}{(x \ - \ 3)·(x \ + \ 2)} \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ = \ 4 \ \ . \ ] $$

Each of the original ratios is represented by a "shifted" rectangular hyperbola with vertical asymptotes at $ \ x \ = \ 3 \ $ and $ \ x \ = \ -2 \ \ . \ $ The upper "branch" of $ \ y \ = \ \frac{k \ - \ 9x}{x \ + \ 2} \ $ is just tangent to the lower branch of $ \ y \ = \ \frac{x \ - \ 1}{x \ - \ 3} \ $ for $ \ k \ = \ 14 \ \ , \ $ and to its upper branch for $ \ k \ = \ 54 \ \ . \ $ (The lower branch of $ \ y \ = \ \frac{k \ - \ 9x}{x \ + \ 2} \ $ never intersects the other curve for $ \ k \ > \ 0 \ \ . \ ) $

The other possibility that may not come to mind easily is that the numerator of $ \ \frac{k \ - \ 9x}{x \ + \ 2} \ $ becomes an exact multiple of the denominator for $ \ k \ = \ -18 \ \ , \ $ so that the original equation "collapses" to $ \ \frac{x \ - \ 1}{x \ - \ 3} \ = \ -9 $ $ \Rightarrow \ x - 1 \ = \ -9x + 27 \ \Rightarrow \ x \ = \ \frac{14}{5} \ \ . \ $ For our quadratic polynomial ratio, we find

$$ \ \mathbf{k \ = \ -18 \ \ :} \quad \quad \frac{10·x^2 \ - \ 8·x \ - \ 56}{(x \ - \ 3)·(x \ + \ 2)} \ \ = \frac{2·(x \ + \ 2)·(5x \ - \ 14)}{(x \ - \ 3)·(x \ + \ 2)} \ \ = \ \ 0 $$ $$ \Rightarrow \ \ x \ = \ -2 \ \ , \ \ \frac{14}{5} \ \ . $$

ADDENDUM: We can obtain the value for $ \ k \ $ directly by dividing $ \ 10·x^2 \ - \ (k \ + \ 26)·x \ + \ (3k \ - \ 2) \ $ by $ \ x + 2 \ \ , \ $ which yields $ \ 10x - (k + 46) + \frac{5·(k \ + \ 18)}{x \ + \ 2} \ \ . \ $ The quotient is a simple linear function only for $ \ k \ = \ -18 \ \ . \ $ There is no comparable result for division by $ \ x - 3 \ \ . $

This produces a single solution owing to the exclusion of $ \ x \ = \ -2 \ $ as permissible. In the geometrical representation, the curve for $ \ y \ = \ \frac{k \ - \ 9x}{x \ + \ 2} \ $ is reduced to the line $ \ y \ = \ -9 \ \ , \ $ which intersects the lower branch of $ \ y \ = \ \frac{x \ - \ 1}{x \ - \ 3} \ $ just once at $ \ \left(\frac{14}{5} \ , \ -9 \right) \ $ and is completely "below" its upper branch.