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A) Exactly one real root in $(-1,0]$

B) Exactly one real root in $[2,5]$

C) Distinct roots greater than 1

D) Distinct roots less than 5

$$x=\frac{4\pm \sqrt {16+4a^2}}{2}$$

$$x=2\pm \sqrt{4+a^2}$$

Also $a\in [-2,2]$

So $a^2\in [0,4]$

Then $x=4,0$ and $x=2(1\pm \sqrt 2)$

Which account for the extrema and minima

These 4 values satisfy A, C and D, but not B. However, the given answer is all 4. How is B correct?

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Aditya
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2 Answers2

1

As you point out $-2 \le a \le -2$ so.

$\sqrt{4 + 0} \le\sqrt{4 + a^2} \le \sqrt {4+4}$

So $2 \le 2+ \sqrt{4 + a^2}\le 2 + \sqrt 8$

So $2 \le 2 + \sqrt{4+a^2} \le 2 + \sqrt 8 < 2 + \sqrt 9 = 2+3 =5$.

So the root $2 + \sqrt{4+a^2} \in [2, 5)\subset [2,5]$.

fleablood
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  • What about $x=4$ when $a=0$? – Aditya May 30 '20 at 16:50
  • Well, What about it? – fleablood May 30 '20 at 16:50
  • It lies inside $[2,5]$, so that gives 2 values – Aditya May 30 '20 at 16:51
  • If $a = 0$ then $2 + \sqrt {4+a^2} = 4 \in [2,5]$ so .....? – fleablood May 30 '20 at 16:51
  • No... your two roots are $2 + \sqrt 4$ and $2 - \sqrt 4$. And $2-\sqrt 4 = 0 \not \in [2,5]$. – fleablood May 30 '20 at 16:52
  • That makes $x=2, 2+2\sqrt 2$ inside $[2,5]$ – Aditya May 30 '20 at 16:52
  • $2+\sqrt 4 \in [2,5]$ – Aditya May 30 '20 at 16:54
  • If $a =0$ then $x = 0$ or $4$. $4$ is in $[2,5]$ and $0$ is not. .... I don't see your confusion. How did you get $x =2,2 + 2\sqrt 2$. That doesn't happen unless $-\sqrt{4 +a^2} = 0$ (which is impossible) at the same time that $+\sqrt{4 + a^2} = 2\sqrt 2$ those can't both happen at the same time. ANd one of them cant ever happen at all. – fleablood May 30 '20 at 16:57
  • $2 + \sqrt 4 \in [2, 5]$ and $2 - \sqrt 4 \in (-1,0]$. What's the problem? – fleablood May 30 '20 at 16:57
  • If $x= 2$ as a root is impossible. If $a =0$ the two roots are $0$ and $4$. – fleablood May 30 '20 at 17:06
  • Actually the positive root is $2 + \sqrt {4+a^2}\in [4,2+2\sqrt 2] \subsetneq [2,5]$. And the non positive root is $2 - \sqrt{4+a^2} \in [-(2\sqrt 2- 2),0]\subsetneq (-1,0]$. That will be true for all $a$. – fleablood May 30 '20 at 17:09
  • So the problem was that the solution cannot be for $a^2=0$ and $a^2=4$ at the same time. It has to be either of those two. Am I getting it right? – Aditya May 31 '20 at 05:18
  • Why are you making this so hard??? $a^2$ can be any value between $0$ and $4$. It could be $0$. It could be $4$ it could be $2.73195$. And the two roots are $2 + \sqrt{4 + a^2}$ and $2- \sqrt{4+a^2}$. The root $2+\sqrt{4+a^2}$ is at least as big as $2+\sqrt{4+ 0} = 4$ but a most as big as $2 + \sqrt{4+4}=2+2\sqrt 2$ (which is less than $5$). And the other root $2-\sqrt{4+a^2}$ at least $2-\sqrt{4+4} = 2-2\sqrt 2$ (which is between $-1$ and $0$) and at most $2-\sqrt{4+0}= 0$... And so one root $2-\sqrt{4+a^2}\in (-1,0]$ no matter what $a$ is. And the other root $2+\sqrt{4+a^2}\in [4,5)$ – fleablood May 31 '20 at 05:40
  • Yes that’s what I said. $a$ can have only one value, while I was counting solutions for both $a^2=0$ and $a^2=4$. There is one solution in $[2,5]$ for every value of $a^2\in [0,4]$. Am I getting this right? – Aditya May 31 '20 at 05:44
0

Since $\sqrt {4+a^2}>0, \forall a\in[-2,2] $ so the equation always have 2 distinct roots. We shall look at the root $2+\sqrt{4+a^2}$.

We know that $\sqrt x$ is an increasing function .

Hence $0\le a^2 \le 4\implies 0< 4+a^2\le 8<9\implies 0<\sqrt {4+a^2}<3\implies 2<2+\sqrt{4+a^2}<5$

Noob mathematician
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