91Ó°ÊÓ

(First, we divide both sides of the equation by the coefficient of $$x^2$$, which is 5. This gives us: $$ x^2 + \frac{17}{5}x + \frac{1}{5} = 0 $$ Now, we will complete the square by finding the square of half of the coefficient of x, then adding and subtracting this value to the equation: $$ \left(\frac{\frac{17}{5}}{2}\right)^2 = \left(\frac{17}{10}\right)^2 = \frac{289}{100} $$ Adding and subtracting $$\frac{289}{100}$$, we get: $$ x^2 + \frac{17}{5}x + \frac{289}{100} - \frac{289}{100} + \frac{1}{5}=0 $$ This can be rewritten as: $$ \left(x + \frac{17}{10}\right)^2 = \frac{289}{100} - \frac{1}{5} $$ We now solve for x: $$ \left(x + \frac{17}{10}\right)^2 = \frac{284}{100} $$ Taking the square root of both sides: $$ x + \frac{17}{10} = \pm \frac{\sqrt{284}}{10} $$ Finally, we subtract $$\frac{17}{10}$$ from both sides: $$ x = \frac{-17 \pm \sqrt{284}}{10} $$ These are the two possible solutions for x using the completing the square method.

(To solve using the quadratic formula, we use the formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ In our equation $$5x^2 + 17x + 1 = 0$$, we have $$a = 5$$, $$b = 17$$, and $$c = 1$$. Substituting the values into the formula, we get: $$ x = \frac{-17 \pm \sqrt{(17)^2 - 4(5)(1)}}{2(5)} $$ Simplifying the expression, we obtain: $$ x = \frac{-17 \pm \sqrt{289 - 20}}{10} $$ $$ x = \frac{-17 \pm \sqrt{269}}{10} $$ These are the two possible solutions for x using the quadratic formula. Comparing with the completing the square method, we have found the same two solutions.