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If $f(x)=4x^2+ax+a-3$ is negative for at least one negative $x$ find all possible values of $a$

I don't know how to find all possible values.

I tried making the lower of the two roots as negative. That does not seem to work out. As $4>0$ the curve will be concave up. There can be numerous cases possible.

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Having distinct real roots requires a positive discriminant, i.e.

$$a^2-16(a-3)=(a-4)(a-12)>0,$$ or$$\color{green}{a<4\lor a>12}.$$

Then, having at least one negative root requires that either the sum or the product of the roots be negative, i.e.

$$-a<0\lor a-3<0,$$which is always true and doesn't restrict the solution set !

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  • $\begingroup$ You give a very nice argument why one of the roots is negative, +1. $\endgroup$ – wythagoras Jun 30 '15 at 10:24
  • $\begingroup$ Yep, it avoids computing the roots. Cheers. $\endgroup$ – Yves Daoust Jun 30 '15 at 10:25
  • $\begingroup$ @YvesDaoust It should be $a<0\lor a-3<0$ $\endgroup$ – Bhaskar Vashishth Jun 30 '15 at 10:33
  • $\begingroup$ Ooops, $-a<0\lor a-3<0$. Fixed. $\endgroup$ – Yves Daoust Jun 30 '15 at 10:39
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You don't have to solve for the roots (which leads to irrational inequations!) to answer.

Indeed, if $f(x)<0$ for at least one $x$, it has two real roots.The discriminant is $\;\Delta=a^2-16a+48=(a-4)(a-12)$, hence it has two real roots if $a<4$ or if $a>12$.

Suppose $\Delta>0$; we'll determine the sign of the roots: $\;f(0)=a-3$, hence:

  • if $a<3$, $0$ separates the roots, i.e. there is one negative and one positive root.
  • if $a>3$, $0$ is outside of the interval of the roots, i.e. they're both positive or both negative. In that case, their sign is also the sign of their half-sum, which is $-\dfrac a8$. Thus both are negative.
  • if $a=3$, the roots are $0$ and $-\dfrac34$.

In all cases, $f(x)$ has at least one negative root, so that $a$ can take any real value in $\,]-\infty,4[\:\cup\:]12,+\infty[$.

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  • $\begingroup$ I think you miscalculated the discriminant... $\endgroup$ – MathGod Jun 30 '15 at 9:30
  • $\begingroup$ @Ishu: thank you for pointing the error. I've updated my answer. Hope there's no more problems :o) $\endgroup$ – Bernard Jun 30 '15 at 9:54
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Your polynomial is $p(x)=4x^2+ax+a-3$ its discriminant is $\Delta=a^2-16a+48$ and its roots are $\frac{-a\pm \sqrt{\Delta}}{8}$. Now there is a $x$ for which value is negative, and it is a parabola opening upwards, tells us that both roots are real and thus $\Delta>0 \implies a^2-16a+48=(a-4)(a-12)>0$ which happens iff $a\in (-\infty,4)\cup(12,\infty)$.

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  • $\begingroup$ Where do you express the condition that there is a negative root ? $\endgroup$ – Yves Daoust Jun 30 '15 at 10:18
  • $\begingroup$ @YvesDaoust I do not think it is a necessary condition for the question that value is negative for at least one negative $x$, just being negative for any value of $x$, gives the same solution. $\endgroup$ – Bhaskar Vashishth Jun 30 '15 at 10:23
  • $\begingroup$ Indeed when there are real roots one is always negative, but you must prove it ! $\endgroup$ – Yves Daoust Jun 30 '15 at 10:28
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The function will be positive when $x$ is very large positive or very large negative. This means that there are two solutions, to your equation, of which at least one is negative.

The smallest root is:

$$\frac{-a -\sqrt{a^2-16(a-3)}}{8}$$

So we want to know when $a^2-16(a-3) > 0$, or $a^2-16a+48 > 0$, or $(a-8)^2-16 > 0$, or $(a-8)^2 > 16$, or $$a-8 > 4 \vee a-8 < -4$$

$$a > 12 \vee a < 4$$

When $a$ is positive, both $-a$ and $-a -\sqrt{a^2-16(a-3)}$ are negative, so then the smallest root is negative.

When $a$ is negative, $$\sqrt{a^2-16a+48}=\sqrt{(a-6)^2+12}>\sqrt{(a-6)^2}=|a-6|=6-a$$

With the the second because of the increasing nature of $\sqrt{x}$, the fourth because $a$ and thus $a-6$ negative is. Therefore we have $$\frac{-a -\sqrt{a^2-16(a-3)}}{8}<\frac{-a - (6-a)}{8}=\frac{-6}{8}=-0.75$$

Thus the solution is

$$a > 12 \vee a < 4$$

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  • $\begingroup$ The inequalities needn't be strict. $\endgroup$ – Yves Daoust Jun 30 '15 at 10:19
  • $\begingroup$ @YvesDaoust Yes they do other wise there is exactly one root and is the polynomial only non-positive, it is never negative. $\endgroup$ – wythagoras Jun 30 '15 at 10:20
  • $\begingroup$ You are right, sorry, I didn't see that the function needed to be strictly negative, just looked at the roots. $\endgroup$ – Yves Daoust Jun 30 '15 at 10:22
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Examining the extrema of $f(x)=4x^2+ax+a-3$ which is a family of upwards open parabolas: $$ f'(x) = 8 x + a = 0 \Rightarrow x = -\frac{1}{8}a \\ f''(x) = 8 > 0 $$ So we have a local minimum at $x = -(1/8)a$. There the function has the value: $$ f(-(1/8)a) = \frac{1}{16}a^2 - \frac{1}{8}a^2 + a - 3 = -\frac{1}{16}a^2 + a - 3 $$ The minima regarding different $a$ values lie on the curve $$ m(x) = -4x^2-8x -3 $$

parabolas (Large image version)

From $$ 0 = -\frac{1}{16} a^2 + a - 3 \Rightarrow \\ a^2 = 16 a - 48 \Rightarrow \\ (a - 8)^2 = 64 - 48 = 16 \Rightarrow \\ a = 8 \pm 4 $$

we get that the minimum is non-negative for $a \in [4, 12]$ and there we will find no negative value for $f$ at all.

So we have a chance to find negative $f$ values for negative argument only for $a \in I = \mathbb{R} \setminus [4,12]$, where the minimum occurs for a negative function value.

For positive $a\in I$ the minimum is at $x_m = -(1/8) a < 0$, so those $a$ are valid.
For negative $a \in I$ for $x = -1/2 < 0$ we have $f(-1/2) = 1 - a/2 + a - 3 = a/2 - 2 < 0$, so those $a$ are valid as well.

This gives $I = \mathbb{R} \setminus [4,12]$ as set of all feasible $a$.

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The graph is a parabola that opens upward. So there are two cases:

  1. $f(0) < 0$

  2. f has a minimum at $x_0 < 0$ such that $f(x_0) < 0$

In the first case, $f(0) = a -3 < 0 \text{ gives } a<3$

In the second case, $f'(x) = 8x + a$ which has a minimum at $x_0 = -\frac{a}{8}$ where we require $a > 0$. Then $f(x_0) = \frac{a^2}{16} - \frac{a^2}{8} + a - 3 = \frac{-a^2 + 16a - 48}{16} = g(a)$ which is a parabola in a that opens downward, and has roots $a = \frac{-16 \pm \sqrt{256-192}}{-2} = 8 \pm 4$ so $g(a)$ is negative for all $0 < a \le 4$ and $a > 12$.

Putting everything together:

$ a \in (-\infty,4) \lor a \in (12,+\infty)$

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You can treat this problem with roots indeed, for as soon as you have two distincts roots $x_1$ and $x_2$, all values between the two roots give a negative $f(x)$.

This you can demonstrate easily.

So to have distinct roots, you need to have a strictly positive discriminant, that is

$a^2-4a+12>0$

You also want to have a negative root, that is $-a +\sqrt{a^2-4a+12}<0$

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The given expression is a quadratic in $x$ with the leading coefficient $\geq 0$. Since, it is given to be negative for at least one $x$, the conditions are:

  1. Discriminant $\geq 0$, that is $a^2-4a+12>0$
  2. At least one root is negative. Solve for the roots and make the smaller one less than $0$.
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