Question

what is the positive solution to the equation $0 = -x^2 + 2x + 1$?
quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
\\(\circ\\) $-2 + \sqrt{2}$
\\(\circ\\) $2 - \sqrt{2}$
\\(\circ\\) $1 + \sqrt{2}$
\\(\circ\\) $-1 + \sqrt{2}$

Explanation:

Step1: Rewrite the equation

The given equation is \(0 = -x^2 + 2x + 1\). Multiply both sides by -1 to make the coefficient of \(x^2\) positive: \(x^2 - 2x - 1 = 0\). Now, for the quadratic formula \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), we have \(a = 1\), \(b=-2\), \(c=-1\).

Step2: Apply the quadratic formula

Substitute \(a = 1\), \(b=-2\), \(c=-1\) into the quadratic formula:

$$
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$$

Step3: Find the positive solution

We have two solutions: \(x = 1+\sqrt{2}\) and \(x = 1 - \sqrt{2}\). Since \(\sqrt{2}\approx1.414\), \(1-\sqrt{2}\approx1 - 1.414=-0.414\) (negative), and \(1+\sqrt{2}\approx1 + 1.414 = 2.414\) (positive). So the positive solution is \(1+\sqrt{2}\).

Answer:

\(1 + \sqrt{2}\) (corresponding to the option "1 + √2")