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

Question

Eva Adamson · Accepted Answer

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

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

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Nathan Axel · Answer

undefined To solve the equation $x^2 - 5 = 0$ graphically, you plot the function $y = x^2 - 5$ on a coordinate plane and find the points where the graph intersects the x-axis. Those points are the solutions to the equation. Algebraically, you can solve it by adding 5 to both sides to isolate $x^2$, then taking the square root of both sides. Keep in mind that there will be two solutions, one positive and one negative. So the solutions are $x = \sqrt{5}$ and $x = -\sqrt{5}$.