Answer :
To solve the quadratic equation [tex]\( x^2 + 2x + 7 = 21 \)[/tex], we need to follow these steps:
1. Rearrange the equation to standard form: The given equation is [tex]\( x^2 + 2x + 7 = 21 \)[/tex]. To bring it to the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], subtract 21 from both sides:
[tex]\[ x^2 + 2x + 7 - 21 = 0 \][/tex]
Simplifying, we get:
[tex]\[ x^2 + 2x - 14 = 0 \][/tex]
2. Identify the coefficients: From the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], we have:
[tex]\[ a = 1, \quad b = 2, \quad c = -14 \][/tex]
3. Calculate the discriminant: The discriminant [tex]\(\Delta\)[/tex] of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the coefficients, we get:
[tex]\[ \Delta = 2^2 - 4 \cdot 1 \cdot (-14) = 4 + 56 = 60 \][/tex]
4. Determine the solutions: The quadratic formula for the roots [tex]\( x \)[/tex] of the equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the given values:
[tex]\[ x = \frac{-2 \pm \sqrt{60}}{2 \cdot 1} \][/tex]
Simplifying further:
[tex]\[ x = \frac{-2 \pm \sqrt{60}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{60}}{2} \][/tex]
Solving, we get two roots [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]:
[tex]\[ x_1 = \frac{-2 + \sqrt{60}}{2}, \quad x_2 = \frac{-2 - \sqrt{60}}{2} \][/tex]
5. Determine the number of positive solutions:
[tex]\[ \sqrt{60} \approx 7.75 \][/tex]
For [tex]\( x_1 \)[/tex]:
[tex]\[ x_1 = \frac{-2 + 7.75}{2} \approx \frac{5.75}{2} \approx 2.87 \][/tex]
For [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = \frac{-2 - 7.75}{2} \approx \frac{-9.75}{2} \approx -4.87 \][/tex]
Since [tex]\( x_1 \)[/tex] is positive and [tex]\( x_2 \)[/tex] is negative, there is only 1 positive solution.
6. Round the greatest solution to the nearest hundredth: The greatest solution is [tex]\( x_1 \)[/tex], which is approximately 2.87.
So, the number of positive solutions is [tex]\( \boxed{1} \)[/tex], and the approximate value of the greatest solution, rounded to the nearest hundredth, is [tex]\( \boxed{2.87} \)[/tex].
1. Rearrange the equation to standard form: The given equation is [tex]\( x^2 + 2x + 7 = 21 \)[/tex]. To bring it to the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], subtract 21 from both sides:
[tex]\[ x^2 + 2x + 7 - 21 = 0 \][/tex]
Simplifying, we get:
[tex]\[ x^2 + 2x - 14 = 0 \][/tex]
2. Identify the coefficients: From the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], we have:
[tex]\[ a = 1, \quad b = 2, \quad c = -14 \][/tex]
3. Calculate the discriminant: The discriminant [tex]\(\Delta\)[/tex] of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the coefficients, we get:
[tex]\[ \Delta = 2^2 - 4 \cdot 1 \cdot (-14) = 4 + 56 = 60 \][/tex]
4. Determine the solutions: The quadratic formula for the roots [tex]\( x \)[/tex] of the equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the given values:
[tex]\[ x = \frac{-2 \pm \sqrt{60}}{2 \cdot 1} \][/tex]
Simplifying further:
[tex]\[ x = \frac{-2 \pm \sqrt{60}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{60}}{2} \][/tex]
Solving, we get two roots [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]:
[tex]\[ x_1 = \frac{-2 + \sqrt{60}}{2}, \quad x_2 = \frac{-2 - \sqrt{60}}{2} \][/tex]
5. Determine the number of positive solutions:
[tex]\[ \sqrt{60} \approx 7.75 \][/tex]
For [tex]\( x_1 \)[/tex]:
[tex]\[ x_1 = \frac{-2 + 7.75}{2} \approx \frac{5.75}{2} \approx 2.87 \][/tex]
For [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = \frac{-2 - 7.75}{2} \approx \frac{-9.75}{2} \approx -4.87 \][/tex]
Since [tex]\( x_1 \)[/tex] is positive and [tex]\( x_2 \)[/tex] is negative, there is only 1 positive solution.
6. Round the greatest solution to the nearest hundredth: The greatest solution is [tex]\( x_1 \)[/tex], which is approximately 2.87.
So, the number of positive solutions is [tex]\( \boxed{1} \)[/tex], and the approximate value of the greatest solution, rounded to the nearest hundredth, is [tex]\( \boxed{2.87} \)[/tex].