Find all values ​of $b$ for which the equation has a unique solution.

Question

Find all values ​​of $b$ for which the equation has a unique solution.
$$x-2=\sqrt{2(b-1)x+1}$$
I found the roots of the equation. $x=b+1\pm\sqrt{b^2+2b-2}$
As I understand, there is one solution: $b=1$, then $x=3$. But I don’t understand how to find it?

Answer 1

You need to find for which values of $b$, the equation has only ONE solution. For example, when $b=1$ you found there was only one solution which is $x=3$. For unknown $b$

$$ x-2=\sqrt{2\left(b-1\right)x+1} \Rightarrow \left(x-2\right)^2=2\left(b-1\right)x+1 $$ Hence $x$ must be a solution of $$ x^2-4x+4=2bx-2x+1\Leftrightarrow x^2-2\left(b+1\right)x+3=0 $$ Then $$ \Delta=4\left(b+1\right)^2-12=4\left(b^2+2b-2\right) $$ Can you discuss whether this kind of expression has no, one or two solutions ? What conditions on $b$ do you need on the solution $x$ so it makes sense ? (Think about the original equation containing a $\sqrt{}$.

Answer 2

I suppose what is required is a real solution. Now \begin{align} x-2=\sqrt{2(b-1)x+1}&\iff (x-2)^2=2(b-1)x+1=0\quad\textbf{and}\quad x-2\ge 0 \cr &\iff p(x)\stackrel{\text{def}}{=}x^2-2(b+1)x+3=0\quad\textbf{and}\quad x\ge 2 \end{align} So there's a single root if & only if the reduced discriminant $\Delta'=(b+1)^2-3=b^2+2b-2\ge 0$ and $2$ separates the roots.

These conditions are easy to translate: they mean that

• $b\le -1-\sqrt 3$ or $b\ge -1+\sqrt 3$ for the first condition,
• $p(2)=3-4b<0\iff b>\frac34$ for the second condition.

The latter condition implies $b>-1+\sqrt 3$ Therefore it is this latter condition which ensures there's a unique solution.

Answer 3

Unlike this problem, where it is important to attend to where various straight lines become tangent to a parabola or pass through its vertex, one will be misled here if a parabola is considered. The function $ \ y \ = \ \sqrt{2(b-1)·x + 1} \ \ , $ for which we are to determine intersections with the line $ \ y \ = \ x - 2 \ \ , $ is only the "upper half" of the "horizontal" parabola $ \ y^2 \ = \ 2·(b-1)·x + 1 \ \ , $ so the line will only have one or no intersections with the square-root curve.

This question can be answered entirely in terms of the domain of the square-root function. The critical value of $ \ b \ $ is $ \ b \ = \ 1 \ \ , $ for which the function curve "degenerates" to the horizontal line $ \ y \ = \ 1 \ \ , $ intersecting $ \ y \ = \ x - 2 \ $ at $ x \ = \ 3 \ \ , $ as you found. For $ \ b \ > \ 1 \ \ , $ the parabola "opens to the right" with domain $ \ x \ \ge \ -\frac{1}{2·(b-1)} \ \ . $ The square-root function then always has one intersection with $ \ y \ = \ x - 2 \ $ at the "positive-root" solution that you found, $ \ x \ = \ (b+1) \ + \ \sqrt{b^2+2b-2} \ \ $ because the vertex of the curve only asymptotically approaches $ \ x = 0 \ $ "from the left" as $ \ b \ \rightarrow +\infty \ \ . $

The issue with the other answers posted is that the square-root curve can only have $ \ y \ \ge \ 0 \ \ . $ For $ \ b < 1 \ \ , $ the parabola "opens to the left" ; it will not intersect the line $ \ y \ = \ x - 2 \ $ for $ \ x \ < \ 2 \ \ , $ the line's $ \ x-$intercept. Now writing the function as $ \ y \ = \ \sqrt{1 - 2(1-b)·x } \ \ , $ we find the domain of the square-root function to be $ \ x \ \le \ \frac{1}{2·(1 - b)} \ \ . $ (Here, the vertex approaches $ \ x = 0 \ $ "from the right" asymptotically as $ \ b \ \rightarrow -\infty \ \ . ) \ $ There will be just one intersection of the two curves at $ \ x \ = \ (b+1) \ + \ \sqrt{b^2+2b-2} \ \ , $ until $ \ b \ $ is reduced to the value at which the vertex of the square-root curve passes "to the left" of the line's $ \ x-$ intercept. So there is no intersection (and no solution to the given equation) for $ \ \frac{1}{2·(1 - b)} \ < \ 2 \ \Rightarrow \ b \ < \ \frac34 \ \ . $ The equation as stated thus has a unique solution for $ \ \mathbf{b \ \ge \ \frac34} \ \ . $

That this must be the intention of the problem posed is that the complete parabola will have two intersections for all values of $ \ b \ $ greater than that at which the parabola just becomes tangent to the line $ \ y \ = \ x - 2 \ \ , $ which occurs for $ \ b \ = \ \sqrt3 - 1 \ \ . $ For $ \ b \ $ less than this value, there are no intersections, as has been discussed. (This problem serves as an illustration of the hazard of creating "spurious" solutions by "squaring the equation." Understanding the geometry of the curves makes it clearer how the solutions are to be interpreted.)

The difference between this answer and the others is illustrated by the graph below; the blue curve is the square-root function for $ \ b \ = \ \frac34 \ \ , $ while the red curve is the parabola for $ \ b \ = \ \sqrt3 - 1 \ \ . $