How do you solve #\sqrt { x ^ { 2} - 2x + 4} = \sqrt { x ^ { 2} + 5x - 12}#?

3 Answers
Jul 21, 2017

Answer:

#x=2 2/7#

Explanation:

Step 1) Square both sides

#sqrt(x^2-2x+4)=sqrt(x^2+5x-12)#

#x^2-2x+4=x^2+5x-12#

Step 2) Put the variables on the left side and the real numbers on the right

#-2x+4=+5x-12#

#-2x-5x=-12-4#

#-7x=-16#

Step 3) Solve

#x= (-16)/-7#

#x=2 2/7#

Jul 21, 2017

Answer:

#x=16/7=2 2/7#

Explanation:

#sqrt(x^2-2x+4)=sqrt(x^2+5x-12)#

Squaring the two sides

#x^2-2x+4=x^2+5x-12#

i.e. #5x+2x=4+12#

or #7x=16#

i.e. #x=16/7=2 2/7#

Jul 21, 2017

Answer:

#x=16/7#

Explanation:

As there s a square root on each sidez we just square each side to remove the square root:
#(sqrt(x^2-2x+4))^2=(sqrt(x^2+5x-12))^2#

#x^2-2x+4=x^2+5x-12#

As each side has one positive #x^2#, we can just remove them by subtracting #x^2#:

#x^2-2x+4-color(red)(x^2)=x^2+5x-12-color(red)(x^2)#

#-2x+4=5x-12#

Now, we can put the #x#s onto the RHS by adding #2x#:

#-2x+4+color(red)(2x)=5x-12+color(red)(2x)#

#4=7x-12#

Now we just need the integers on the LHS by adding #12#:

#4+color(red)(12)=7x-12+color(red)(12)#

#16=7x#

To get #x# on its own, we just divide both sides by 7:

#16/color(red)(7)=(cancel(7)x)/(cancel(color(red)(7)))#

#x=16/7#