How do you solve #x^2-5=0# graphically and algebraically?

1 Answer
Mar 20, 2016

Answer:

#x=+-sqrt(5) # exactly, or approximately #+-2.236# to 3 decimal places

Explanation:

Given:#" " x^2=5=0#

To solve mathematically we need to have just 1 #x# and for it to be on one side of the equals sign and everything else on the other side.

To get the x term on its own.

Add #color(blue)(5) # to both sides of the equation

#" "color(brown)(x^2-5color(blue)(+5)=0color(blue)(+5)#

#x^2+0=5#

#x^2=5#

But #x^2# is the same as #x xx x# and we need to have just 1 not 2 of them.

Take the square root of both sides

#sqrt(x^2)=sqrt(5)#

So #x=sqrt(5)" " # which is 2.236 to 3 decimal places

But both #(-2.236 )xx(-2.236 ) = 5" "#

and#" "(+2.236 )xx(+2.236 )=5#

So #x=+-sqrt(5) # exactly or approximately #+-2.236# to 3 decimal places

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B