Question

Which of the following statements are true about the equation below?

Answer 1

Answer:

Option 1, 2, 5 are the correct answer.

Explanation:

 We have the quadratic equation, $x^2-6x+2=0$

 First derivative of the equation is given by 2x - 6 = 0

               So x = 3

At x = 3 the value of quadratic equation is extreme, corresponding y is given by $3^2-6*3+2=11-18=-7$, So extreme value is at (3,-7)

Second derivative of the quadratic equation is given by 2 ( positive value)

 Second derivative is positive so graph of equation has a minimum value.

Now root of the equation $x^2-6x+2=0$ is given by

            $\frac{6+\sqrt{(-6)^2-4*1*2}} {2} =3+\sqrt{7}$

                                             or

            $\frac{6-\sqrt{(-6)^2-4*1*2}} {2} =3-\sqrt{7}$

Option 1, 2, 5 are the correct answer.              

             

Answer 2

The answer are

The graph of the quadratic equation has a minimum value.

The extreme value is at the point (3,-7).

The solutions are x = 3 ± $\sqrt{7}$.

Here is how it looks on Plato or any multiple select.

Hope this helps