Answer
(a) $x=3 \pm \sqrt 5$
(b) $3$
(c) The average of the solutions is equal to $x-$ coordinate of the vertex.
Work Step by Step
(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.
