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Math  /  Algebra

Question7.2 Determine the real solution(s) (if they exist) of the following quadratic equations (a) x2+2x+2=0x^{2}+2 x+2=0 (b) 9x2−12x−21=09 x^{2}-12 x-21=0 (c) 4x2−20x+25=04 x^{2}-20 x+25=0 (+) x2=14x−50x^{2}=14 x-50 (+) On August 7th, the temperature in Klagenfurt was 15 degree Celsius at 6 am and 29 degree Celsius at 1 pm . What was the average temperature gradient (in degree Celsius per hour) during that morning?

Studdy Solution

STEP 1

1. We are solving quadratic equations, which may have real or complex solutions.
2. The quadratic formula can be used to find the solutions of quadratic equations.
3. The temperature gradient is calculated as the change in temperature divided by the change in time.

STEP 2

1. Solve quadratic equation (a) using the quadratic formula.
2. Solve quadratic equation (b) using the quadratic formula.
3. Solve quadratic equation (c) using the quadratic formula.
4. Solve quadratic equation (+) by rearranging into standard form and using the quadratic formula.
5. Calculate the average temperature gradient.

STEP 3

For equation (a) x2+2x+2=0 x^2 + 2x + 2 = 0 , identify coefficients: a=1 a = 1 , b=2 b = 2 , c=2 c = 2 . Use the quadratic formula:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Substitute the values:
x=−2±22−4⋅1⋅22⋅1 x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} x=−2±4−82 x = \frac{-2 \pm \sqrt{4 - 8}}{2} x=−2±−42 x = \frac{-2 \pm \sqrt{-4}}{2}
Since the discriminant b2−4ac=−4 b^2 - 4ac = -4 is negative, there are no real solutions. The solutions are complex.

STEP 4

For equation (b) 9x2−12x−21=0 9x^2 - 12x - 21 = 0 , identify coefficients: a=9 a = 9 , b=−12 b = -12 , c=−21 c = -21 . Use the quadratic formula:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Substitute the values:
x=−(−12)±(−12)2−4⋅9⋅(−21)2⋅9 x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 9 \cdot (-21)}}{2 \cdot 9} x=12±144+75618 x = \frac{12 \pm \sqrt{144 + 756}}{18} x=12±90018 x = \frac{12 \pm \sqrt{900}}{18} x=12±3018 x = \frac{12 \pm 30}{18}
Calculate the solutions:
x=12+3018=4218=73 x = \frac{12 + 30}{18} = \frac{42}{18} = \frac{7}{3} x=12−3018=−1818=−1 x = \frac{12 - 30}{18} = \frac{-18}{18} = -1
The real solutions are x=73 x = \frac{7}{3} and x=−1 x = -1 .

STEP 5

For equation (c) 4x2−20x+25=0 4x^2 - 20x + 25 = 0 , identify coefficients: a=4 a = 4 , b=−20 b = -20 , c=25 c = 25 . Use the quadratic formula:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Substitute the values:
x=−(−20)±(−20)2−4⋅4⋅252⋅4 x = \frac{-(-20) \pm \sqrt{(-20)^2 - 4 \cdot 4 \cdot 25}}{2 \cdot 4} x=20±400−4008 x = \frac{20 \pm \sqrt{400 - 400}}{8} x=20±08 x = \frac{20 \pm \sqrt{0}}{8} x=208=52 x = \frac{20}{8} = \frac{5}{2}
There is one real solution, x=52 x = \frac{5}{2} .

STEP 6

For equation (+) x2=14x−50 x^2 = 14x - 50 , rearrange into standard form:
x2−14x+50=0 x^2 - 14x + 50 = 0
Identify coefficients: a=1 a = 1 , b=−14 b = -14 , c=50 c = 50 . Use the quadratic formula:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Substitute the values:
x=−(−14)±(−14)2−4⋅1⋅502⋅1 x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4 \cdot 1 \cdot 50}}{2 \cdot 1} x=14±196−2002 x = \frac{14 \pm \sqrt{196 - 200}}{2} x=14±−42 x = \frac{14 \pm \sqrt{-4}}{2}
Since the discriminant b2−4ac=−4 b^2 - 4ac = -4 is negative, there are no real solutions. The solutions are complex.

STEP 7

Calculate the average temperature gradient. The temperature at 6 am is 15°C and at 1 pm is 29°C. The time difference is 7 hours (from 6 am to 1 pm).
Temperature change = 29−15=14 29 - 15 = 14 degrees Celsius.
Average temperature gradient = Temperature changeTime change=147=2 \frac{\text{Temperature change}}{\text{Time change}} = \frac{14}{7} = 2 degrees Celsius per hour.
The real solutions for the quadratic equations are: (a) No real solutions. (b) x=73 x = \frac{7}{3} and x=−1 x = -1 . (c) x=52 x = \frac{5}{2} . (+) No real solutions.
The average temperature gradient is 2 2 degrees Celsius per hour.

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