2

$$x^4-5x^2+4>0$$

Let $x^2=a$

Then $a^2-5a+4>0$

Solving the inequation we get $$a \in (-\infty , 1)\cup (4, \infty)$$

So now $x^2<1$

Then $x<1$ and $x>-1$

Therefore $x\in (-1,1)$

And $x^2>4$

So $x>2$ and $x<-2$

Which implies $x\in (-\infty, -2)\cup (2,\infty)$

Taking the intersection of both values does not yield any result, however the answer given is $$x\in (-\infty, -2)\cup (-1,1) \cup (2,\infty)$$ which can be obtained by taking their union. But since we have to take intersection, where am I going wrong?

Toby Mak
  • 16,827
Aditya
  • 6,191
  • There is one wrongly used and in your solution that leads to an obstacle in your understanding: it should be $x<-2$ or $x>2.$ Each of these possibilities satisfies, you do not need to do the intersection. – user376343 Sep 07 '19 at 11:33

3 Answers3

5

You don't have to take intersection, take union. Because your solution for $a$ says $$a<1 \ \ \ or\ \ \ a>4$$. Now, you have found values of $x$ such that $x^2<1$ and values of $x$ such that $x^2>4$, so you should take union now.

A more direct approach would be to factorise completely in one go, as follows.

$$x^4-5x^2+4=(x^2-1)(x^2-4)$$ $$=(x+2)(x+1)(x-1)(x-2)$$

Now apply wavy curve method.

Hope it helps:)

Martund
  • 14,706
  • 2
  • 13
  • 30
  • I know what you mean by 'wavy curve method', but I guess you would need a bit of explanation. Maybe you can suggest sketching the curve and marking out the roots? – Toby Mak Sep 07 '19 at 11:34
  • 1
    I am expecting OP to know wavy curve method, since he used it to find solution for $a$. – Martund Sep 07 '19 at 11:36
  • Yeah, although factorizing Bi quadratic isn’t the easiest thing in the world, So I use this instead. – Aditya Sep 07 '19 at 11:38
  • But don’t we take intersection? I have solved many questions like this where taking the intersection got me the right answer. – Aditya Sep 07 '19 at 12:25
  • There is no rule which is applicable in all situations. You have to judge the situation and then decide. Here you are getting or in solution of $a$, so you should take union for corresponding solutions of $x$. – Martund Sep 07 '19 at 12:38
1

Hint.

$x^2>4 \Rightarrow (x>2) \lor (x\lt -2)$

AgentS
  • 12,195
0

Hint: Your inequality above can be written as $$\left(x^2-\frac{5}{2}\right)^2-\frac{9}{4}>0 $$ Now use the formula $$a^2-b^2=(a-b)(a+b)$$

Dr. Sonnhard Graubner
  • 95,283