Questions: Consider this quadratic equation. [ x^2+2 x+7=21 ] The number of positive solutions to this equation is . The approximate value of the greatest solution to the equation, rounded to the nearest hundredth, is .

Solution Steps

To solve the given quadratic equation \(x^2 + 2x + 7 = 21\), we first need to bring it to the standard form \(ax^2 + bx + c = 0\). Then, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the solutions. The discriminant \(b^2 - 4ac\) will help us determine the number of real solutions. If the discriminant is positive, there are two real solutions; if zero, one real solution; and if negative, no real solutions. Finally, we calculate the solutions and determine the greatest one.

We start with the quadratic equation given in the problem: \[ x^2 + 2x + 7 = 21 \] Rearranging this equation to standard form gives us: \[ x^2 + 2x - 14 = 0 \]

Next, we calculate the discriminant \(D\) using the formula: \[ D = b^2 - 4ac \] Substituting the values \(a = 1\), \(b = 2\), and \(c = -14\): \[ D = 2^2 - 4 \cdot 1 \cdot (-14) = 4 + 56 = 60 \]

Since the discriminant \(D = 60\) is positive, this indicates that there are two distinct real solutions to the equation.

Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{D}}{2a} \] we find the solutions: \[ x = \frac{-2 \pm \sqrt{60}}{2 \cdot 1} = \frac{-2 \pm \sqrt{60}}{2} \] Simplifying further: \[ x = -1 \pm \frac{\sqrt{60}}{2} \] Calculating \(\sqrt{60}\): \[ \sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15} \] Thus, the solutions become: \[ x = -1 \pm \sqrt{15} \]

The two solutions are: \[ x_1 = -1 + \sqrt{15} \quad \text{and} \quad x_2 = -1 - \sqrt{15} \] Since \(-1 - \sqrt{15}\) is negative, the only positive solution is: \[ x_1 = -1 + \sqrt{15} \]

To find the approximate value of the greatest solution, we calculate: \[ \sqrt{15} \approx 3.872983 \] Thus: \[ x_1 \approx -1 + 3.872983 \approx 2.872983 \] Rounding to the nearest hundredth gives: \[ x_1 \approx 2.87 \]

Final Answer