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Find the permissible value(s) of $k$ given in options below for which both roots of the equation $4x^2 -2x+k = 0$ are completely in $(-1,1)$ .

The options given are : $-1,0,2$ and $-3$ .

By completing the square, we have

$$(2x -\frac{1}{2})^2 = -k + \frac{1}{4}$$

Also, $$x \in (-1,1) \implies (2x -\frac{1}{2})^2 \in (0, \frac{25}{4})$$

Hence, $$-k + \frac{1}{4} \in (0, \frac{25}{4})\implies k\in(-6,-\frac{1}{4})$$

Hence, from the options, I got that $k$ can be equal to $-1,0,-3$ . However, the answer key says that answers are $-1$ and $0$ only. Where did I go wrong ?

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    $\begingroup$ Are you sure about the implication $x\in (-1,1)\Rightarrow \left(2x-1/2\right)^2\in (9/4,25/4)$? What if $x=1/4$? $\endgroup$
    – Kandinskij
    May 8, 2023 at 10:45
  • $\begingroup$ @Kandinskij Oops sorry, it should be $0$ in place of $9/4$ . But still, It will not match the answer key. I'll edit it immediately. Thanks for pointing. $\endgroup$ May 8, 2023 at 10:47
  • $\begingroup$ Okay, now are you sure that the INVERSE of this implication holds? $\endgroup$
    – Kandinskij
    May 8, 2023 at 10:49
  • $\begingroup$ @Kandinskij Since according to question, roots should be in $(-1,1)$ , so wouldn't a complete range of $k$ would be derived by taking Inverse implication ? $\endgroup$ May 8, 2023 at 10:52
  • $\begingroup$ Why not use the quadratic equation solver formula? $\endgroup$
    – NoChance
    May 8, 2023 at 11:07

4 Answers 4

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If you compute the two roots of you polynomial (by using the quadratic formula) you'll get: $$x=\frac{2\pm\sqrt{4-16k}}{8}=\frac{1\pm\sqrt{1-4k}}{4}$$ Moreover, since you want to have real roots you need the condition $1-4k\geq0\Rightarrow k\leq \frac{1}{4}$.

Now, observe that $\frac{1-\sqrt{1-4k}}{4}\leq \frac{1+\sqrt{1-4k}}{4}$ so you have to solve two inequalities: $$-1<\frac{1-\sqrt{1-4k}}{4}$$$$\frac{1+\sqrt{1-4k}}{4}<1$$ The first inequality gives you $k>-6$ and the second one gives you $k>-2$. Hence, the solution is $-2<k\leq\frac{1}{4}$. In particular $k=-1,0$ works. As @user0 pointed out in a comment, this inequalities are easily solved, you just have to be careful about flipping the inequality when multiplying by a $-$ sign.

The problem with your argument is that you have shown that $x\in(-1,1)\Rightarrow k\in(-6,\frac{1}{4}]$, which is completely right since $-2<k\leq \frac{1}{4}$. However, the converse is not proved, so you don't know which of the values of $(-6,\frac{1}{4}]$ works.

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  • $\begingroup$ Oh I get it. The reason that I'm getting more values is because squaring always increases the number of solutions, right? Thanks ! $\endgroup$ May 8, 2023 at 11:19
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    $\begingroup$ @Dstarred Just a typo, thanks $\endgroup$
    – Marcos
    May 8, 2023 at 11:22
  • $\begingroup$ @An_Elephant exactly, you have to be careful with squaring equations. $\endgroup$
    – Marcos
    May 8, 2023 at 11:22
  • $\begingroup$ agreed, you have considered the entire interval, sometimes these inequalities can be best be solved by plotting on a simple number line and marking a "common" region $\endgroup$ May 8, 2023 at 11:24
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So we have, $p(x) = 4x^2 -2x +k=0$ with roots in $(-1,1)$

As the above quadratic has both roots within $(-1,1)$ or in other words the two roots lie between $X_1$ and $X_2$. The conditions that a quadratic $P(x)=ax^2+bx+c$ would satisfy are;

(i) $\Delta$ ≥ 0

(ii) $X_1$ < $\frac{-b}{2a}$ < $X_2$

(iii) $ap(X_1) > 0$ and $ap(X_2) > 0$

And your answer should be $0$ and $-1$.

So when you apply the first one, you get $\frac{1}{4} ≥ k$

when you apply the third one, you get $k>-6$ and $k>-2$ so the common interval for k is $(-2,\frac{1}{4}]$.

NOTE- $X_1$ and $X_2$ here is -1 and 1 respectively, and ap( $X_1$) > 0 here means for a quadratic $p(x)=ax^2+bx+c$, ap( $X_1$) would be a(a$( X_1)^2$+b( $X_1$)+c)

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  • $\begingroup$ @User0 you should improve your notation. If you dont write $X_1=-1$ and $X_2=1$ how am I supposed to deduce it? Moreover $ap(X_1)$ is also a bad notation, since you already know that in the problem $a=4$ and you have not said what $a$ was. $\endgroup$
    – Marcos
    May 8, 2023 at 11:25
  • $\begingroup$ @User0 don't want to be mean. Just I think you can improve a bit your notation and will help us to understand you better. $\endgroup$
    – Marcos
    May 8, 2023 at 11:26
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    $\begingroup$ Im still adapting to this website but thanks for your advice, also as i know @An_Elephant, we both can understand such notations used in my country $\endgroup$ May 8, 2023 at 11:27
  • $\begingroup$ @User0 Its ok, I kow mathjax is difficult for beginers and it is very easy to think that some notation is standard, but it is not used outside some countries. But you'll be ok with a bit of extra practice in proof writing. $\endgroup$
    – Marcos
    May 8, 2023 at 11:30
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As you've already found, $ \ 4x^2 -2x + k \ = \ 4·(x -\frac14)^2 \ + \ \left(k - \frac14 \right) \ \ , $ so this polynomial is represented by an "upward-opening" parabola with its vertex at $ \ \left( \ \frac14 \ , \ k - \frac14 \ \right) \ \ . \ $ The polynomial cannot then have real zeroes $ \ (x-$intercepts of the parabola) for $ \ k \ > \ \frac14 \ \ . \ $ For $ \ k \ = \ \frac14 \ \ , \ $ there is a "double zero" ("touching intercept") at $ \ x \ = \ \frac14 \ \ . $

As $ \ k \ $ is decreased, the vertex will "shift downward" and the zeroes/intercepts will "spread outward" symmetrically from $ \ x \ = \ \frac14 \ \ . \ $ [We can easily check that at $ \ k \ = \ 0 \ \ , \ $ we have $ \ 4x^2 - 2x \ = \ 2x·(2x - 1) \ \ , \ $ placing the zeroes at $ \ x = \ 0 \ $ and $ \ x \ = \ \frac12 \ \ . \ ] \ $ The "critical value for $ \ k \ $ will be that at which the larger zero reaches $ \ x \ = \ 1 \ \ , \ $ which we can determine from $$ 4·1^2 \ - \ 2·1 \ + \ k_c \ \ = \ \ 0 \ \ \Rightarrow \ \ k_c \ \ = \ \ -2 \ \ . \ $$ "Symmetrical spreading" indicates that the other zero lies at $ \ x \ = \ \frac14 - \left(1 - \frac14 \right) \ = \ -\frac12 \ \ . \ $ We can check this by factoring $ \ 4x^2 - 2x - 2 \ = \ 2·(2x^2 - x - 1) \ = \ 2·(2x + 1)·(x - 1) \ = \ 0 \ \ . \ $

Hence, the zeroes for this polynomial are both in the open interval $ \ (-1 \ , \ 1) \ $ for $ \ \frac14 \ \ge \ k \ > \ -2 \ \ . \ $ Among the available choices, this only permits $ \ 0 \ $ and $ \ -1 \ \ . $

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  • $\begingroup$ Nice ! I too had thought of graphical solution but not in this way. $\endgroup$ May 10, 2023 at 9:03
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First of all, $p(x)$ is continuous everywhere as it is polynomial. So, applying intermediate value theorem.

Put $0$ in $x$ and decide what k should be. Also put $-1$ in $x$ to have $p(0).p(-1) \le 0$.

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  • $\begingroup$ $p(0)$ cannot be determined as positive or negative. $p(0)$ is positive if both roots are of same sign and $p(0)$ is negative if both roots are of opposite signs. $\endgroup$ May 8, 2023 at 11:27

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