For what $k$, are the roots of $p(x) = 4x^2 -2x +k=0$ in $(-1,1)$?

Question

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 ?

Answer 1

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.

Answer 2

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)

Answer 3

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 $p(0).p(-1) \le 0$.

Answer 4

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 \ \ . $