Question

select all the correct answers.
which of the following statements are true about the equation below?
$x^2 - 6x + 2 = 0$

• the extreme value is at the point $(3, -7)$.
• the solutions are $x = -3 \pm \sqrt{7}$.
• the extreme value is at the point $(7, -3)$.
• the graph of the quadratic equation has a maximum value.
• the solutions are $x = 3 \pm \sqrt{7}$.
• the graph of the quadratic equation has a minimum value.

Explanation:

Step1: Determine the vertex (extreme value)

For \(ax^2 + bx + c = 0\), vertex x-coordinate is \(-\frac{b}{2a}\). Here \(a=1\), \(b=-6\), so \(x = -\frac{-6}{2(1)} = 3\). Substitute \(x=3\) into \(x^2 -6x +2\): \(3^2 -6(3)+2 = 9-18+2=-7\). Vertex is (3,-7).

Step2: Solve the quadratic equation

Using quadratic formula \(x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}\). \(b^2 -4ac = (-6)^2 -4(1)(2)=36-8=28\), so \(x = \frac{6 \pm \sqrt{28}}{2} = \frac{6 \pm 2\sqrt{7}}{2}=3 \pm \sqrt{7}\).

Step3: Determine if it's minimum or maximum

Since \(a=1>0\), the parabola opens upwards, so it has a minimum value.

Answer:

The extreme value is at the point (3, -7), The solutions are \(x = 3 \pm \sqrt{7}\), The graph of the quadratic equation has a minimum value.