(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: .
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)