Question

Find the real solutions, if any, of each equation. Use any method. $$ x^{2}+\sqrt{2} x=\frac{1}{2} $$

Short answer

The solutions are \( x = 1 - \frac{\sqrt{2}}{2} \) and \( x = -1 - \frac{\sqrt{2}}{2} \).

Step-by-step solution

Move all terms to one side of the equation

Rewrite the given equation by moving all terms to the left side to set the equation to 0: \[ x^2 + \sqrt{2}x - \frac{1}{2} = 0 \]

Identify coefficients for the quadratic formula

Identify the coefficients in the quadratic equation, where the general form is \[ ax^2 + bx + c = 0 \].Here, we have:\[a = 1\] \[b = \sqrt{2}\] \[c = -\frac{1}{2}\]

Apply the quadratic formula

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), substitute the values identified in Step 2:\[ x = \frac{-\sqrt{2} \pm \sqrt{(\sqrt{2})^2 - 4(1)(-\frac{1}{2})}}{2(1)}\]

Simplify under the square root

Simplify the expression under the square root, (\[ \sqrt{2})^2 = 2 \], and\[ 4(1)(-\frac{1}{2}) = -2 \], resulting in \[ x = \frac{-\sqrt{2} \pm \sqrt{2 + 2}}{2}\]

Simplify further

Simplify the expression inside the square root and calculate the values:\[ x = \frac{-\sqrt{2} \pm \sqrt{4}}{2} = \frac{-\sqrt{2} \pm 2}{2}\]

Solve for x

Finally, compute the two possible solutions for x:\[ x = \frac{-\sqrt{2} + 2}{2} \] and \[ x = \frac{-\sqrt{2} - 2}{2} \]Which simplifies to:\[ x = 1 - \frac{\sqrt{2}}{2} \] and \[ x = -1 - \frac{\sqrt{2}}{2} \]

Key concepts

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \[ax^2 + bx + c = 0\]. This formula allows us to find the values of \(x\) that make the equation true. The quadratic formula is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
Here's a simple breakdown:

• \(a\) is the coefficient of \(x^2\)
• \(b\) is the coefficient of \(x\)
• \(c\) is the constant term

By inserting these coefficients into the formula, we can solve for \(x\).
In our example: \((a = 1, b = \sqrt{2}, c = -\frac{1}{2})\), we substitute these into the quadratic formula.
This results in: \[x = \frac{-\sqrt{2} \pm \sqrt{(\sqrt{2})^2 - 4(1)(-\frac{1}{2})}}{2(1)}\]. This step sets us up for simplification and finding the real solutions.

real solutions

After substituting the coefficients into the quadratic formula, the next step involves simplifying the expression under the square root, known as the discriminant. The discriminant, \(b^2 - 4ac\), determines the nature of the solutions:

• If the discriminant is positive, there are two distinct real solutions.
• If the discriminant is zero, there is exactly one real solution.
• If the discriminant is negative, there are no real solutions (the solutions are complex).

In the given problem, the discriminant is:\[b^2 - 4ac = (\sqrt{2})^2 - 4(1)(-\frac{1}{2}) = 2 + 2 = 4\]. Since this value is positive, we know there will be two real solutions.
These real solutions occur where the quadratic formula divides the overall value into two outcomes: \[x = \frac{-\sqrt{2} + 2}{2}\] and \[x = \frac{-\sqrt{2} - 2}{2}\].

simplifying expressions

Simplifying expressions is crucial in finding the final solutions to quadratic equations. Here's a useful approach on how to do this:

• First, combine like terms inside the square root.
• Next, simplify the numerator separately for each solution.
• Finally, reduce the fraction, if possible.

In our exercise: \[x = \frac{-\sqrt{2} \pm \sqrt{4}}{2} = \frac{-\sqrt{2} \pm 2}{2}\]. We break this into two solutions:
\[x = \frac{-\sqrt{2} + 2}{2}\] and \[x = \frac{-\sqrt{2} - 2}{2}\].
Simplifying these further, we get:
\[x = 1 - \frac{\sqrt{2}}{2}\] and \[x = -1 - \frac{\sqrt{2}}{2}\].
Breaking down each part and ensuring clarity can make solving quadratic equations less intimidating. By systematically applying the quadratic formula, identifying real solutions, and simplifying expressions, you can effectively manage quadratic equations of various complexities.