Answer:
[tex]x=4[/tex]
Step-by-step explanation:
Given triangle is a right angle triangle
To find out x we use Pythagorean theorem
[tex]c^2=a^2+b^2[/tex]
where c is the hypotenuse
c=square root (20),[tex]a=x and b=x-2[/tex]
Plug in the values in the formula
[tex](\sqrt{20} )^2= x^2+(x-2)^2[/tex]
[tex]20=x^2+x^2-4x+4[/tex]
[tex]20=2x^2-4x+4[/tex]
Subtract 20 on both sides
[tex]0=2x^2-4x+-16[/tex]
Divide whole equation by 2
0=x^2-2x-8
factor the right hand side
[tex]0=(x-4)(x+2)[/tex]
Set each factor =0 and solve for x
[tex]x-4=0[/tex] so [tex]x=4[/tex]
[tex]x+2=0[/tex] so [tex]x=-2[/tex]
The length x cannot be negative, so [tex]x=4[/tex]
