solve the inequality
$x^2-(5a^2-a+15)x+6a^4+30a^2+54

Question

solve the inequality
$x^2-(5a^2-a+15)x+6a^4+30a^2+54<0$
where a is constant

akashdeepdeb · Answer

I think you have to split the middle term!
Something I am trying Right now! :D

ujjwal · Answer

I know.. Split middle term so that the product of the obtained terms after splitting is equal to 6a^4+30a^2+54
But then, i am not getting anything.. :\

akashdeepdeb · Answer

Try splitting
6a^4 as 3a^2 + 2a^2
54 as 6 and 9
30a^2 as 6a^2 - 5a^2 then maybe it can work!
Hold on let me try!

akashdeepdeb · Answer

Did you try doing that and did you get it?

ujjwal · Answer

i didn't get anything.. unfortunately..
while solving a similar problem previously, i had factorized the last term (which in this case is 6a^4+30a^2+54) and that had worked..

akashdeepdeb · Answer

Bahhh!
Hold on...

akashdeepdeb · Answer

I'll msg you the answer! :D

anonymous · Answer

i think that u have a quadratic in terms of x with a|dw:1377507140731:dw| leading coefficient equal to +1... so the interval which satisfies the inequality will be $x_1<x<x_2$ where $x_1$ and $x_2$ are the roots of quadratic

anonymous · Answer

how about using direct formula for finding roots ?? just an idea

akashdeepdeb · Answer

O.O
I tried splitting the middle term!

dls · Answer

$$x^2-(5a^2-a+15)x+6a^4+30a^2+54<0$$
$(5a^2-a+15$) is always +ve since D<0
Hence,
Co-efficient of x >0
Also, $6a^4+30a^2+54$ is always positive since even power.
Therefore,
$$x^2-(Sum~of~many~positive~terms)<0$$
How about this? o.O logical method :o

akashdeepdeb · Answer

Answer?

ujjwal · Answer

I don't have that.

akashdeepdeb · Answer

Not you!
@DLS  ?

anonymous · Answer

$$x^2-(5a^2-a+15)x+6a^4+30a^2+54=0$$has the roots$$x_1=3(a^2-a+3)$$$$x_2=2(a^2+a+3)$$

ujjwal · Answer

@mukushla how did you get those roots?

akashdeepdeb · Answer

Excellent! :D

anonymous · Answer

well direct formula$$x_1,x_2=\frac{-b\pm \sqrt{\Delta}}{2a}$$

akashdeepdeb · Answer

Quadratic formula right?

akashdeepdeb · Answer

Hhah...Direct method! O.O

dape · Answer

We can get more information, so the problem is not finished. We can use the roots of $x$ to factorize the whole expression:
$$ x^2-(5a^2-a+15)x+6a^4+30a^2+54=(x-3(a^2-a+3))(x-2(a^2+a+3))<0 $$
Now, think about what happens when we fix a value of $a$, depending on what it is, the signs of the two factors will change, but we will always need that one is positive and one is negative. So for different values of $a$, we get different conditions for $x$. This would be one way to get the full solution.

ujjwal · Answer

well, what i think is we have 3 cases.. 1) when $3(a^2-a+3)>2(a^2+a+3)$, we have, $3(a^2-a+3)>x>2(a^2+a+3)$ 2)when $3(a^2-a+3)=2(a^2+a+3)$, there is no solution 3)when $3(a^2-a+3)<2(a^2+a+3)$, then, $3(a^2-a+3)

dape · Answer

That's right! Or that's what I got. You can also simplify the inequalities for $a$, you should then get 3 disconnected solution sets.

ujjwal · Answer

can't simplify inequalities for a.. I tried.. without success..

dape · Answer

So you see that the a inequalities all have the same form, if we rearrange and combine, we get a polynomial the roots of which (where it switches sign) gives us the different solution sets, the polynomial is
$$ a^2-5a+3 $$
And the roots are $a=(5\pm\sqrt{13})/2$.

dape · Answer

For your solution 1, for example, when $a>(5+\sqrt{13})/2$, the first inequality is satisfied. But it is also satisfied for $a<(5-\sqrt{13})/2$. So we have 2 solutions for what looked like the first solution. This is because the polynomial is positive in these two regions.

dape · Answer

Now, for your 3) or second solution, you want the region of a where the polynomial is negative. This is the third and final solution set.