Question

How do you solve x²-10x+9=0 by completing the square?

Answer 1

$x^2-10x+9=0 \\ \\x^2-2\cdot5x+5^2-5^2+9=0\\ \\(x-5)^2-25+9=0\\ \\(x-5)^2=16\\ \\x-5=4\ \ \vee\ \ x-5=-4\\ \\x=9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=1$

Answer 2

$x^2-10x+9=0\\\\\underbrace{x^2-2x\cdot5+5^2}_{(*)}-5^2+9=0\\\\(x-5)^2-25+9=0\\\\(x-5)^2-16=0\ \ \ /+16\\\\(x-5)^2=16\iff x-5=-\sqrt{16}\ \vee\ x-5=\sqrt{16}\\\\x-5=-4\ \vee\ x-5=4\\\\x=-4+5\ \vee\ x=4+5\\\\x=1\ \vee\ x=9$



$(*)\ (a-b)^2=a^2-2ab+b^2$