Question

Using quadratic formula find the roots of the equation \(2x^2 - 2\sqrt{2}x + 1 = 0\).

Hint:

When the discriminant \(D=0\), the quadratic expression is a perfect square. In this case, \(2x^2 - 2\sqrt{2}x + 1 = 0\) can be written as \((\sqrt{2}x - 1)^2 = 0\), which directly gives \(\sqrt{2}x = 1\) or \(x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\).

Answer 1

Step 1: Understanding the Concept:
The quadratic formula is a general solution for finding the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\).

Step 2: Key Formula or Approach:
The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] The term \(D = b^2 - 4ac\) is the discriminant, which tells us the nature of the roots.

Step 3: Detailed Explanation:
The given equation is \(2x^2 - 2\sqrt{2}x + 1 = 0\).
Comparing this to the standard form \(ax^2 + bx + c = 0\), we have: \(a = 2\), \(b = -2\sqrt{2}\), \(c = 1\).
First, let's calculate the discriminant \(D\): \[ D = b^2 - 4ac = (-2\sqrt{2})^2 - 4(2)(1) \] \[ D = (4 \times 2) - 8 = 8 - 8 = 0 \] Since the discriminant is 0, the equation has two real and equal roots.
Now, apply the quadratic formula: \[ x = \frac{-(-2\sqrt{2}) \pm \sqrt{0}}{2(2)} \] \[ x = \frac{2\sqrt{2}}{4} \] Simplify the fraction: \[ x = \frac{\sqrt{2}}{2} \] Since the roots are equal, both roots are \(\frac{\sqrt{2}}{2}\).

Step 4: Final Answer:
The roots of the equation are \(x = \frac{\sqrt{2}}{2}\) and \(x = \frac{\sqrt{2}}{2}\).