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Given: (1) $x+\sqrt{a^2-x^2}=b$, $(a,b,x)\subset \mathbb R$, $a>0$, $b>0$.

Find: number of roots for (1), given possible values for $a$ and $b$.

This is a question from a book for the preparation for math contests.

It states as final answer: (a) 1 root if $b<a$; and (b) 2 roots if $a<b<a\sqrt{2}$.

I'm having difficulties on finding this answer. I don't know whether it is correct or perhaps I'm not finding the right approach.

I started moving $x$ in (1) to the left, to get $$\sqrt{a^2-x^2}=b-x$$ Before proceeding with squaring both sides, I saved 2 needed conditions for checking the final solution (c1) $a^2-x^2\ge 0$ and (c2) $b-x\ge 0$. Then squaring both sides, we get: $$a^2-x^2=b^2+x^2-2bx\Leftrightarrow 2x^2-2bx+(b^2-a^2)=0$$ with discriminant $\triangle$ defined by: $$\triangle=4(2a^2-b^2)$$ From this it is easy to see that a condition for 2 roots is (c3) $\sqrt{2}a>b,$ and for 1 root is (c4) $\sqrt{2}a=b,$ as $a>0$ and $b>0$. Then I find the roots as $$x=\frac{2b\pm \sqrt{\triangle}}{4}=\frac{b\pm \sqrt{2a^2-b^2}}{2}$$ From this point, I can't see a way to reach the stated answer, if it is right.

Full solutions or helpful hints are welcome. Sorry if it is a duplicate.

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  • $\begingroup$ The problem only depends on a single parameter: posing $x=as$ and $b=a\beta$, the equation depends only on $\beta$ which simplifies the discussion. $\endgroup$ – Paul Enta Nov 11 '17 at 11:24
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As $a>0$, posing $x=as$ and $b=\beta a$, the problem is equivalent to discuss the real roots of $$s+\sqrt{1-s^2}=\beta$$ with $\beta>0$. The function has real values for $\left| s\right|<1$. Moreover, $f(s)=s+\sqrt{1-s^2}$ has a single maximum at $s=1/\sqrt{2}$ with $f(1/\sqrt{2})=\sqrt{2}$, $f(0)=f(1)=1$ and finally $f(-1)<0$. Thus, there are no real root for $\beta>\sqrt{2}$, 2 real roots for $1<\beta<\sqrt{2}$, a single real root for $0<\beta<1$ and a double root at $\beta=\sqrt{2}$.

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    $\begingroup$ For $a=2$, $b=1$, $\beta =1/2$ there is one solution $\endgroup$ – Lozenges Nov 11 '17 at 12:13
  • $\begingroup$ You are right. I should have notice that the equation has no solution for $\left| s\right|>1$. I edit the answer accordingly. $\endgroup$ – Paul Enta Nov 11 '17 at 12:41
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As you wrote, we have $$2x^2-2bx+b^2-a^2=0$$ with$$a^2-x^2\ge 0\quad\text{and}\quad b-x\ge 0,$$ i.e. $$-a\le x\le a\quad\text{and}\quad x\le b\tag1$$

Here, let us separate it into cases :

  • Case 1 : If $(a\sqrt 2\gt)\ a\gt b$, then $(1)\iff -a\le x\le b$. We have $\frac{b+\sqrt{2a^2-b^2}}{2}\gt \frac{b+\sqrt{2b^2-b^2}}{2}=b$ and $b\gt 0\gt \frac{b-\sqrt{2a^2-b^2}}{2}\gt \frac{b-2a}{2}=\frac b2-a\gt -a$. So, in this case, 1 root.

  • Case 2 : If $a\le b\lt a\sqrt 2$, then $(1)\iff -a\le x\le a$, and $$\small\frac{b+\sqrt{2a^2-b^2}}{2}\le a\iff \sqrt{2a^2-b^2}\le 2a-b\iff 2a^2-b^2\le (2a-b)^2\iff (a-b)^2\ge 0$$which indeed holds, and $$a\gt \frac{b+\sqrt{2a^2-b^2}}{2}\gt\frac{b-\sqrt{2a^2-b^2}}{2}\gt \frac{b-\sqrt{2b^2-b^2}}{2}=0$$So, in this case, 2 roots.

  • Case 3 : If $b=a\sqrt 2$, then $(1)\iff -a\le x\le a$, and $x=\frac{a}{\sqrt 2}$ satisfies this. So, in this case, 1 root.

  • Case 4 : If $b\gt a\sqrt 2$, the equation has no real solutions.

Therefore,

1 root if $a\gt b$ or $b=a\sqrt 2$

2 roots if $a\le b\lt a\sqrt 2$

0 root if $b\gt a\sqrt 2$

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You just forgot to check those conditions you mentioned. Indeed, you should check that for what values of $a$ and $b$ the conditions $-a\le x\le a$ and $b-x \ge 0$ are satisfied. The difficulty is to consider a few if then conditions.

If $b=\sqrt{2}a$ then $x=\frac{b}{2}$ is a potential solution. We should make sure that this solution satisfies our inequalities, that is, we should have $-a\le\frac{b}{2}\le a$ and $b-\frac{b}{2}\ge0$ which is equivalent to $b\le 2a$. But $b=\sqrt{2}a$ so this leads to $\sqrt{2}\le 2$ which is identically true. So if $b=\sqrt{2}a$ then the only answer is $x=\frac{b}{2}=\frac{\sqrt{2}}{2}a$.

If $b\gt\sqrt{2}a$ then $\Delta:=\sqrt{2a^2-b^2}<0$ and there is not real solution.

If $b\lt\sqrt{2}a$ then two potential answers will be $x_1=\frac{b}{2}+\frac{\sqrt{2a^2-b^2}}{2}$ and $x_2=\frac{b}{2}-\frac{\sqrt{2a^2-b^2}}{2}$. I guess you can see what to do from here. Just put $x_1$ and $x_2$ into the inqualities and see what restrictions will be obtained on $a$ and $b$. Then check these restrictions with the condition $b\lt\sqrt{2}a$ to see under what conditions $x_1$ and $x_2$ can be a solution.

Below, you can see a plot of the function $f(x)=x+\sqrt{a^2-x^2}$ with $a=1$. You can see easily that for what values of $b$ the equation $f(x)=b$ has a solution and indeed exactly how many solutions.

enter image description here

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You want to study the function $$ f(x)=x+\sqrt{a^2-x^2} $$ defined over $[-a,a]$. We have $f(-a)=-a$, $f(a)=a$; moreover $$ f'(x)=1-\frac{x}{\sqrt{a^2-x^2}} $$ for $x\in(-a,a)$. The derivative can only vanish where $$ x=\sqrt{a^2-x^2} $$ so $x\ge0$ and $x^2=a^2-x^2$, that is, $x=a/\sqrt{2}$. Note that $$ f(a/\sqrt{2})=\frac{a}{\sqrt{2}}+\frac{a}{\sqrt{2}}=a\sqrt{2} $$ and that $a/\sqrt{2}$ is a point of absolute maximum for $f$.

Therefore the equation $x+\sqrt{a^2-x^2}=b$ has

  • no solution for $b<-a$
  • one solution for $-a\le b\le a$
  • two solutions for $a<b<a\sqrt{2}$
  • one solution for $b=a\sqrt{2}$
  • no solutions for $b>a\sqrt{2}$

An algebraic solution.

The equation $\sqrt{a^2-x^2}=b-x$ has solutions only for $x\le b$. Then you can square: $a^2-x^2=b^2-2bx+x^2$ or $2x^2-2bx+b^2-a^2$. The roots of this equation are $$ \frac{b-\sqrt{2a^2-b^2}}{2} \qquad\text{and}\qquad \frac{b+\sqrt{2a^2-b^2}}{2} $$ There is no solution for $2a^2-b^2<0$, that is, $b>a\sqrt{2}$.

There will be two solutions when $b<a\sqrt{2}$ and $$ \frac{b+\sqrt{2a^2-b^2}}{2}\le b $$ that is, $$ \sqrt{2a^2-b^2}\le b $$ or $b\ge a$. Hence, $a\le b<a\sqrt{2}$.

There will be one solution when the largest root is greater than $b$, that is $b<a$, but also the smallest root is $\le b$, that is $$ \frac{b-\sqrt{2a^2-b^2}}{2}\le b $$ which is always satisfied when $b>0$.

When the discriminant is $0$, then there is a single solution.

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It is fairly easy when done graphically. Roots must suffice the equation $\sqrt{a^2-x^2} = x - b$ Notice that $$ y=\sqrt{a^2-x^2}$$ is equation of upper half of the circle of radius $a$ (square it remebering $y\geq0$). On the right hand side $$y=b-x$$ is then just a line intersecting vertical axis at $y$ descending with slope equal to $-1$. Here's an example plotted with WolframAlpha

enter image description here

It then breaks down to discussion on number of intersections of these two graphs:

They intersect once when distance from line to $(0,0)$ is equal to radius a $a$. Then we have $b^2 = 2a^2$ from Pythagorean theorem, so $b=a\sqrt{2}$ guarantee existance of exactly one real root.

For $b>a\sqrt{2}$ there are clearly no intersections.

Proceeding with $b<a\sqrt{2}$ graphs start to have twice common points, but only to the point when $b=a$ (the intersections are then exactly at $(a,0)$ and $(0,a)$).

For $b<a$ we again have one intersection in the upper left quarter. We would have $0$ roots for $b<-a$ also, however we restrict $b$ to be greater than $0$, so that fully solves the problem. Concluding:

  • $0$ for $b>a\sqrt{2}$
  • $1$ for $b \in \{b>0:b=a\sqrt{2} \lor b< a\}$ or
  • $2$ for $a\leq b<a\sqrt{2}$

Plus with this approach you avoid as much algebra as possible. In exchange it may be not so versatile, but still it's always worth to give it a try with similar problems.

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it must be $$a\geq |x|$$ and $$b\geq x$$ after squaring we get the equation $$0=2x^2-2bx+b^2-a^2=0$$ solving this equation we get $$x_1=\frac{b}{2}+\frac{1}{2}\sqrt{2a^2-b^2}$$ $$x_2=\frac{b}{2}-\frac{1}{2}\sqrt{2a^2-b^2}$$ so we get the following solution set $$x=b$$ and $a=b$ $$x=\frac{1}{2}(b\pm i\sqrt{b^2-2a^2}$$ and $$0<a<\frac{b}{\sqrt{2}}$$ $$x=\frac{b}{2}$$ and $$a=\frac{b}{\sqrt{2}}$$ $$x=\frac{1}{2}(b\pm \sqrt{2a^2-b^2})$$ and $$\frac{b}{\sqrt{2}}<a<b$$ $$x=\frac{1}{2}(b-\sqrt{2a^2-b^2})$$ and $$a\geq b$$ Observe that i need time nto type all these formulas

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  • $\begingroup$ and thank you very much for the $-1$ you just started, when i was typing $\endgroup$ – Dr. Sonnhard Graubner Nov 11 '17 at 11:35

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