Answer: [tex]x=4[/tex]
Step-by-step explanation:
The Pythagorean Theorem is:
[tex]a^2=b^2+c^2[/tex]
Where the hypotenuse is "a" and "b" and "c" are the legs of the triangle.
Knowing that:
[tex]a=\sqrt{20}\\b=x-2\\c=x[/tex]
Susbstitute values into the Pythagorean Theorem and solve for x:
[tex](\sqrt{20})^2=(x-2)^2+x^2[/tex]
Remember that the square a binomial is:
[tex](a\±b)^2=a^2\±2ab+b^2[/tex]
Then:
[tex]20=x^2-2(x)(2)+2^2+x^2\\20=x^2-4x+4+x^2\\2x^2-4x-16=0[/tex]
Use the Quadratic formula:
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}\\\\x=\frac{-(-4)\±\sqrt{(-4)^2-4(2)(-16)}}{2(2)}[/tex]
[tex]x_1=4\\x_2=-2[/tex]
Choose the positive value.
Then:
[tex]x=4[/tex]
