5.3 practice problem
period ______ date ______
1) use the quadratic formula to solve the equation ( x^2 + 6x - 10 = 0 ).
2) use the quadratic formula to solve ( x^2 + 2x + 2 = 0 )
3) deon throws a ball into the air. the height, ( h ), of the ball, in meters, at time ( t ) seconds is modeled by the function ( h(t) = -5t^2 + t + 4 )
a) when will the ball hit the ground?
b) will the ball reach a height of 5 meters?
4) use the discriminant to identify the number and type of solutions for ( 25x^2 - 20x + 4 = 0 )
5) find the value(s) of ( k ) that will cause the equation to have the given number and type of solutions. ( 5x^2 + kx + 5 = 0 ), 1 real solution.
6) rick claims that the equation ( x^2 + 5x + 9 = 0 ) has no solution. jenny claims that there are two solutions. explain how rick could be correct and explain how jenny could be correct.

Question

5.3 practice problem
period ______ date ______
1) use the quadratic formula to solve the equation ( x^2 + 6x - 10 = 0 ).
2) use the quadratic formula to solve ( x^2 + 2x + 2 = 0 )
3) deon throws a ball into the air. the height, ( h ), of the ball, in meters, at time ( t ) seconds is modeled by the function ( h(t) = -5t^2 + t + 4 )
   a) when will the ball hit the ground?
   b) will the ball reach a height of 5 meters?
4) use the discriminant to identify the number and type of solutions for ( 25x^2 - 20x + 4 = 0 )
5) find the value(s) of ( k ) that will cause the equation to have the given number and type of solutions. ( 5x^2 + kx + 5 = 0 ), 1 real solution.
6) rick claims that the equation ( x^2 + 5x + 9 = 0 ) has no solution. jenny claims that there are two solutions. explain how rick could be correct and explain how jenny could be correct.

Accepted Answer

### Problem 1: Solve $ x^2 + 6x - 10 = 0 $ using quadratic formula
# Explanation:
## Step 1: Identify $ a, b, c $
For $ ax^2 + bx + c = 0 $, here $ a = 1 $, $ b = 6 $, $ c = -10 $.
## Step 2: Quadratic formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
Substitute values: $ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-10)}}{2(1)} $
## Step 3: Simplify discriminant
$ 6^2 - 4(1)(-10) = 36 + 40 = 76 $, so $ x = \frac{-6 \pm \sqrt{76}}{2} $
Simplify $ \sqrt{76} = 2\sqrt{19} $, thus $ x = \frac{-6 \pm 2\sqrt{19}}{2} = -3 \pm \sqrt{19} $
# Answer: $ x = -3 + \sqrt{19} $ or $ x = -3 - \sqrt{19} $

### Problem 2: Solve $ x^2 + 2x + 2 = 0 $ using quadratic formula
# Explanation:
## Step 1: Identify $ a, b, c $
$ a = 1 $, $ b = 2 $, $ c = 2 $
## Step 2: Quadratic formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
Substitute: $ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} $
## Step 3: Simplify discriminant
$ 4 - 8 = -4 $, so $ x = \frac{-2 \pm \sqrt{-4}}{2} $
$ \sqrt{-4} = 2i $, thus $ x = \frac{-2 \pm 2i}{2} = -1 \pm i $
# Answer: $ x = -1 + i $ or $ x = -1 - i $ (complex solutions)

### Problem 3a: When will the ball hit the ground? ($ h(t) = -5t^2 + t + 4 $)
# Explanation:
## Step 1: Set $ h(t) = 0 $
$ -5t^2 + t + 4 = 0 $ (multiply by -1: $ 5t^2 - t - 4 = 0 $)
## Step 2: Identify $ a, b, c $
$ a = 5 $, $ b = -1 $, $ c = -4 $
## Step 3: Quadratic formula $ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
Substitute: $ t = \frac{1 \pm \sqrt{(-1)^2 - 4(5)(-4)}}{2(5)} $
## Step 4: Simplify discriminant
$ 1 + 80 = 81 $, so $ t = \frac{1 \pm 9}{10} $
Two solutions: $ t = \frac{1 + 9}{10} = 1 $ or $ t = \frac{1 - 9}{10} = -0.8 $ (time can't be negative, so $ t = 1 $ second)
# Answer: The ball hits the ground at $ t = 1 $ second.

### Problem 3b: Will the ball reach 5 meters?
# Explanation:
## Step 1: Set $ h(t) = 5 $
$ -5t^2 + t + 4 = 5 $ → $ -5t^2 + t - 1 = 0 $ (multiply by -1: $ 5t^2 - t + 1 = 0 $)
## Step 2: Calculate discriminant $ D = b^2 - 4ac $
$ a = 5 $, $ b = -1 $, $ c = 1 $, so $ D = (-1)^2 - 4(5)(1) = 1 - 20 = -19 $
## Step 3: Analyze discriminant
Since $ D < 0 $, no real solutions (no time $ t $ where height is 5m)
# Answer: No, the ball will not reach 5 meters.

### Problem 4: Discriminant for $ 25x^2 - 20x + 4 = 0 $
# Explanation:
## Step 1: Identify $ a, b, c $
$ a = 25 $, $ b = -20 $, $ c = 4 $
## Step 2: Discriminant formula $ D = b^2 - 4ac $
Substitute: $ D = (-20)^2 - 4(25)(4) $
## Step 3: Simplify
$ 400 - 400 = 0 $, so discriminant is 0 (one real solution, repeated root)
# Answer: Discriminant is 0, so 1 real (repeated) solution.

### Problem 5: Find $ k $ for $ 5x^2 + kx + 5 = 0 $ to have 1 real solution
# Explanation:
## Step 1: 1 real solution means discriminant $ D = 0 $
## Step 2: Discriminant formula $ D = b^2 - 4ac $
For $ 5x^2 + kx + 5 = 0 $, $ a = 5 $, $ b = k $, $ c = 5 $
So $ D = k^2 - 4(5)(5) = k^2 - 100 $
## Step 3: Set $ D = 0 $
$ k^2 - 100 = 0 $ → $ k^2 = 100 $ → $ k = \pm 10 $
# Answer: $ k = 10 $ or $ k = -10 $

### Problem 6: Analyze $ x^2 + 5x + 9 = 0 $
# Brief Explanations:
- **Rick (no solution):** For a quadratic $ ax^2 + bx + c = 0 $, discriminant $ D = b^2 - 4ac $. Here, $ a = 1 $, $ b = 5 $, $ c = 9 $. $ D = 5^2 - 4(1)(9) = 25 - 36 = -11 $. Since $ D < 0 $, no real solutions (complex solutions exist, but if we consider only real numbers, Rick is correct).
- **Jenny (two solutions):** If we consider complex numbers (or the set of all complex numbers), the quadratic formula gives two solutions ($ x = \frac{-5 \pm \sqrt{-11}}{2} $, two distinct complex solutions). So Jenny is correct if we include complex solutions.
# Answer: Rick is correct (real solutions: $ D < 0 $, no real roots); Jenny is correct (complex solutions: two distinct complex roots).

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Question

Explanation:

Problem 1: Solve \( x^2 + 6x - 10 = 0 \) using quadratic formula

Step 1: Identify \( a, b, c \)

Step 2: Quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Step 3: Simplify discriminant

Step 1: Identify \( a, b, c \)

Step 2: Quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Step 3: Simplify discriminant

Step 1: Set \( h(t) = 0 \)

Step 2: Identify \( a, b, c \)

Step 3: Quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Step 4: Simplify discriminant

Answer:

Problem 2: Solve \( x^2 + 2x + 2 = 0 \) using quadratic formula