(4x-3)^2-(2x+5)^2=0 tìm x

Hi

Your question would be better with  "  Could you help me please ? " at the end. Quite sure your mum told you that some time in the past.

I will show you how to solve but you will do the calculus. Math is all about rules, methods and training.  If you dont practice, you will never progress.

rule  :  you must know by heart your remarkable identities.

Here we have  :   ( 4x-3)² - (2x+5)² = 0

It 's of the  form  A² -B² =  ( A+B) ( A-B)  

here A is  :   4x-3   and  B is  2x+5  

Now let's apply rule  :   ( 4x-3)² - (2x+5)² =  ( (4x-3) +( 2x+5) )  ( (4x-3) - (2x+5)

                                                                  =    ( 4x-3 +2x+5 )  ( 4x-3 -2x-5 )

                                                                  =     ( 6x +2)  ( 2x-8)  

so :  ( 4x-3)² - (2x+5)² =  ( 6x +2)  ( 2x-8)  

so  ( 4x-3)² - (2x+5)²  = 0   means  :  ( 6x +2)  ( 2x-8)    = 0

Rule A*B = 0 if A = 0  or B = 0

A  is  6x+2     and B = 2x-8

So I let you now work  :  

You have to solve :   6x+2 = 0

and   2x-8 = 0

Solutions of this two calculus will be the solutions of  ( 4x-3)² - (2x+5)²  = 0  

Have fun !                                    

it has the same slope and a different y-intercept

hope this : )