Thread: inequality

Originally Posted by kastamonu

(3x-2)^2-(x+4)^2<0 find the solution set.

I suggest expanding the two squared terms, simplifying the result, and solving the related equality. You will then know where the squared terms are equal, and so can determine where their difference is negative. Does that make sense?

I tried to find it like that but this is not the answer:
8x^2-20x-12
(4x-4)(2x+3)
(-2/3,1) is the solution set.

Originally Posted by kastamonu

(3x-2)^2-(x+4)^2<0 find the solution set.

$(3x-2)^2-(x+4)^2<0$

$(9x^2-12x+4)-(x^2+8x+16)<0$

$8x^2-20x-12<0$

$4(x-3)(2x+1)<0$

$(x-3)(2x+1)<0$

Now. For this to be true the two factors must be of opposite signs.

So either

$(x-3)<0$ AND $(2x+1)>0$

OR

$(x-3)>0$ AND $(2x+1)<0$

this all reduces to

$x<3$ AND $x>-\dfrac 1 2$

OR

$x>3$ AND $x<-\dfrac 1 2$

the second condition is clearly impossible so we are left with the first which gives

$-\dfrac 1 2 < x < 3$

You made a mistake somewhere.

Hello, kastamonu!

A slightly different approach . . .

We have: . . . . a difference of squares.

. .

. . . .

. . . .


We have a up-opening parabola: .

. . Q: Where is it negative (below the x-axis)?

. . A: between its x-intercepts

Therefore: .

Many Thanks.I made a mistake when I was factorizing,

It should be:
(4x+2)(2x-6)

But I got 2 valuable answers. many Thanks.

I am first finding the roots. Then I am arranging from the largest to the smallest.Later I am checking the signs of the coefficints. I am multiplying the signs.I f it is negative, I am putting a"+" to the right of the largest number and -,+.....

+ -1/2 - 3 +
Solution set is (-1/2,3)