JEE Quadratics

Algebra Level 5

$\large f(x)=ax^2 - (3 + 2a)x + 6, \quad a\ne 0$

If the set of values of $a$ for which $f(x)$ is postive for exactly three distinct negative integral values of $x$ is $(c,d]$ , then find the value of $c^2 + 16d^2$ .

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3 solutions

Chew-Seong Cheong
Aug 17, 2016

From $f(x) = ax^2-(3+2a)x+6 = (ax-3)(x-2)$ , we note that the roots of $f(x)$ are $\dfrac 3a$ and $2$ . The three negative integral roots must come from $\dfrac 3a$ within the range $[-4,-3)$ . That is:

$\begin{aligned} -4 \le & \frac 3a < -3 \\ - \frac 14 \ge & \frac a3 > -\frac 13 \\ - \frac 34 \ge & a > -1 \end{aligned}$

$\implies c = - 1$ , $d = - \frac 34$ and $c^2 + 16d^2 = 1 + 9 = \boxed{10}$

Nice solution. About your note, I think (c,d] should be correct, as at a=c=-1, the negative root is x=-3, so you only get f(-2) and f(-1) positive, which is not good enough. Same reason for d to be inclusive.

Wei Chen - 4 years, 10 months ago

That was my thought $x \in (-4,-3]$ so $a \in [-1,-\frac 34)$ .

Chew-Seong Cheong - 4 years, 10 months ago

Actually x should be in [-4,-3), not including -3. Because f(-3)=0, you only get f(-2) and f(-1) positive, but the problem asks for f(x) to be positive at 3 negative integers. On the other hand, -4 should be inclusive, as f(-4)=0, then you get f(-3),f(-2) and f(-1) to be positive. So the range of a should be (-1,-3/4].

Wei Chen - 4 years, 10 months ago

@Wei Chen – Check the graph that I plotted in the solution.

Chew-Seong Cheong - 4 years, 10 months ago

@Chew-Seong Cheong – Yeah, we can use your graph. If a=-1, then the negative root is 3/a=-3, so f(-3)=0. Since the graph is a parabola opening down, and the other root positive at x=2, you only get f(-2) and f(-1) to be positive, where is the THIRD positive f(x) (at x=negative integer) coming from? can we focus on this question and get this point straight.

Wei Chen - 4 years, 10 months ago

@Wei Chen – You are right. I got confused because it is negative.

Chew-Seong Cheong - 4 years, 10 months ago

Why we take (-4,-3) as range

Himanahu Singh - 4 years, 10 months ago

See the graph above. Three negative integer values of $x$ are needed.

Chew-Seong Cheong - 4 years, 10 months ago

Akshat Sharda
Aug 16, 2016

Notice that $f(2)=0$ , by thinking of the graph of $f(x)$ we can find that the other root will lie between $[-4,-3)$ .

So, by solving $f(-4)≤0$ and $f(-3)>0$ we get $\left(-1,-\frac{3}{4}\right]$ .

$\Rightarrow (-1)^2+16\left(-\frac{3}{4}\right)^2=1+9=\boxed{10}$

Subh Mandal
Oct 2, 2016

f(2)=0 There fore |p| where p is negative root should be at a distance b/w 3,4 from origin ,by solving we get root as 3/a where a is -ve, |3/a|>3,|3/a|<4 solving these 2 we get answer as 10.

JEE Quadratics

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