Algebra 1

(a) $x=3 \pm \sqrt 5$ (b) $3$ (c) The average of the solutions is equal to $x-$ coordinate of the vertex.
(a) Consider the expression: $x^2-6x+4=0$ or, $x^2-6x=-4$ Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=-6$ To complete the square, add $9$ on both sides. $x^2-6x+9=-4+9$ $\implies (x-3)^2=5$ or, $x=3 \pm \sqrt 5$ (b) Consider the expression: $x^2-6x+4=0$ or, $x^2-6x=-4$ Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=-6$ Plug in $x=-\frac{b}{2a}=3$ c) $\frac{(3 + \sqrt 5)+(3 -\sqrt 5)}{2}=3$ Thus, $3=3$ Hence, the average of the solutions is equal to $x-$ coordinate of the vertex.