Certainly! To find the positive solution of the quadratic equation [tex]\(0 = -x^2 + 2x + 1\)[/tex], we will solve it step by step using the quadratic formula. The quadratic formula for an equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In the equation [tex]\(0 = -x^2 + 2x + 1\)[/tex], we identify the coefficients as follows:
- [tex]\(a = -1\)[/tex]
- [tex]\(b = 2\)[/tex]
- [tex]\(c = 1\)[/tex]
Next, we calculate the discriminant ([tex]\(\Delta\)[/tex]), which is given by [tex]\(b^2 - 4ac\)[/tex]:
[tex]\[ \Delta = b^2 - 4ac = 2^2 - 4(-1)(1) = 4 + 4 = 8 \][/tex]
Then, we apply the quadratic formula to find the roots:
[tex]\[ x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-2 \pm \sqrt{8}}{2(-1)} \][/tex]
Now, simplify the expression under the square root:
[tex]\[ \sqrt{8} = 2\sqrt{2} \][/tex]
So our solutions become:
[tex]\[ x_{1,2} = \frac{-2 \pm 2\sqrt{2}}{-2} \][/tex]
Break this into two separate solutions:
[tex]\[ x_1 = \frac{-2 + 2\sqrt{2}}{-2} = 1 - \sqrt{2} \][/tex]
[tex]\[ x_2 = \frac{-2 - 2\sqrt{2}}{-2} = 1 + \sqrt{2} \][/tex]
Among the solutions [tex]\(1 - \sqrt{2}\)[/tex] and [tex]\(1 + \sqrt{2}\)[/tex], we want the positive solution.
Since [tex]\(1 - \sqrt{2}\)[/tex] is a negative number and [tex]\(1 + \sqrt{2}\)[/tex] is a positive number:
[tex]\[ \text{Positive solution} = 1 + \sqrt{2} \][/tex]
Thus, the positive solution to the equation [tex]\(0 = -x^2 + 2x + 1\)[/tex] is:
[tex]\[ 1 + \sqrt{2} \][/tex]
