the values of b for which

touches the x-axis are:

Given:
​$$y = 2x^{2} - bx + 2$$​​
Formula used:
For a quadratic equation to touch the x-axis, its discriminant must be zero.
​$$D = b^{2} - 4ac$$

Alter: 

This is a parabola because the highest power of x is 2.

Since the coefficient of x² is positive (2), the parabola opens upwards.

“Touches the x-axis” means the parabola just touches the x-axis at one point (it does not cut it).

So:

The vertex lies exactly on the x-axis

There is only one real root

​Solution:
Here,
a = 2
b = -b
c = 2
Discriminant:
​$$D = (-b)^{2} - 4(2)(2)\\D = b^{2} - 16$$​​
For touching the x-axis,
​$$b^{2} - 16 = 0\\b^{2} = 16\\b = \pm 4$$​​
The correct answer is (b) $$b = \pm 4.$$

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