If (x^2 + 3)*(x^2 - 1)

Question

If  (\sqrt{x^2} + 3)*(\sqrt{x^2} - 1 ) < 0 ,  \sqrt{x^2} ?

A  -3 < \sqrt{x^2} < 1

B.  0 < \sqrt{x^2} < 1

C.  -1 < \sqrt{x^2} < 3

D.  -3 < \sqrt{x^2} < -1

E.  1 < \sqrt{x^2} < 3

Bunuel · Answer

Can you please tell me the source of this question? Thank you!

Archit3110 · Answer

(\sqrt{x^2} + 3)*(\sqrt{x^2} - 1 ) < 0 
√x^2 = lxl 
lxl will always be + so x cannot be -1<x<0
Option B is correct

KarishmaB · Answer

\sqrt{x^2} = |x|

So we are given
(|x| + 3)*(|x| - 1) < 0

We know that |x| can never be negative. So |x| + 3 will always be positive.

For the product to be negative, (|x| - 1) < 0 i.e. |x| < 1
Hence, 0 <= |x| < 1 (because |x| can be 0 but cannot be negative)

Hence 
 0 <= \sqrt{x^2} < 1

NoHalfMeasures · Answer

Indian MBA entrance exam

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem Series ↧↧↧

Given that \((\sqrt{x^2} + 3)*(\sqrt{x^2} - 1 ) < 0\) and we need to find the value of \(\sqrt{x^2}\)

Let \(\sqrt{x^2}\) = y

=> We have (y + 3) * (y - 1) < 0

Using Sine Wave Method we will plot the two points -3 and 1 and the Sine Wave as shown below

Attachment:

IP-11 image.jpg [ 23.35 KiB | Viewed 1339 times ]

Since we are looking for < 0
So, solution will be -3 < y < 1

=> -3 < \(\sqrt{x^2}\) < 1

So, Answer will be A
Hope it helps!

Watch the following video to MASTER Inequalities

Ghoskrp3 · Answer

Answer should be A!: Let, \sqrt{x^2} = y Then eqn. would become: (y + 3) * (y - 1) < 0 On solving, you get the roots of the equations on 2 points, y = -3 and y = 1 and is negative for any value in between (-3 to 1, table shared below) Value (y=) Equation Findings -4 (y^2)+2y-3 => 5 which is a positive value, > 0 -3 (y^2)+2y-3 => 0, thus still unacceptable -2 (y^2)+2y-3 => -3 < 0, thus acceptable -1 (y^2)+2y-3 => -4 < 0, thus acceptable 0 (y^2)+2y-3 => -3 < 0, thus acceptable 1 (y^2)+2y-3 => 0, thus still unacceptable 2 (y^2)+2y-3 => 5 which is a positive value, > 0 Hence, Answer has to be, A) −3 < \sqrt{x^2} < 1

If (x^2 + 3)*(x^2 - 1) < 0, (x^2)^(1/2) ?

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Value (y=)	Equation	Findings
-4	(y^2)+2y-3	=> 5 which is a positive value, > 0
-3	(y^2)+2y-3	=> 0, thus still unacceptable
-2	(y^2)+2y-3	=> -3 < 0, thus acceptable
-1	(y^2)+2y-3	=> -4 < 0, thus acceptable
0	(y^2)+2y-3	=> -3 < 0, thus acceptable
1	(y^2)+2y-3	=> 0, thus still unacceptable
2	(y^2)+2y-3	=> 5 which is a positive value, > 0