unit 2a - cw2.7
solving - quadratic formula
algebra ii - polynomial functions
page 26
solve for all zeros using the quadratic formula.**must be written in standard form!
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x^2 + 6x + 8 = 0$
$x^2 - 2x - 5 = 0$
$x^2 - 4x - 91 = 7$
$x^2 - 10x + 26 = 8$
$x^2 - 4x + 1 = 5$
$x^2 - 2x = -20$
$x^2 = 18x + 40$
$8x^2 + 16x = 42$

Question

Accepted Answer

Let's solve the first quadratic equation $ x^2 + 6x + 8 = 0 $ using the quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $.

# Explanation:
## Step 1: Identify $ a $, $ b $, $ c $
For the quadratic equation $ ax^2 + bx + c = 0 $, in $ x^2 + 6x + 8 = 0 $, we have $ a = 1 $, $ b = 6 $, $ c = 8 $.

## Step 2: Calculate the discriminant $ D = b^2 - 4ac $
Substitute $ a = 1 $, $ b = 6 $, $ c = 8 $ into the discriminant formula:
$ D=(6)^2 - 4\times1\times8 $
$ D = 36 - 32 $
$ D = 4 $

## Step 3: Apply the quadratic formula
Substitute $ a = 1 $, $ b = 6 $, $ D = 4 $ into the quadratic formula:
$ x=\frac{-6\pm\sqrt{4}}{2\times1} $
Since $ \sqrt{4}=2 $, we have:
$ x=\frac{-6\pm2}{2} $

## Step 4: Find the two solutions
For the plus sign:
$ x=\frac{-6 + 2}{2}=\frac{-4}{2}=-2 $
For the minus sign:
$ x=\frac{-6-2}{2}=\frac{-8}{2}=-4 $

# Answer:
The zeros of the quadratic equation $ x^2 + 6x + 8 = 0 $ are $ x=-2 $ and $ x = - 4 $

We can also solve the second quadratic equation $ x^2-2x - 5=0 $ using the quadratic formula.

# Explanation:
## Step 1: Identify $ a $, $ b $, $ c $
For the quadratic equation $ ax^2+bx + c = 0 $, in $ x^2-2x - 5=0 $, we have $ a = 1 $, $ b=-2 $, $ c=-5 $

## Step 2: Calculate the discriminant $ D=b^2-4ac $
Substitute $ a = 1 $, $ b=-2 $, $ c = - 5 $ into the discriminant formula:
$ D=(-2)^2-4\times1\times(-5) $
$ D = 4 + 20 $
$ D=24 $

## Step 3: Apply the quadratic formula
Substitute $ a = 1 $, $ b=-2 $, $ D = 24 $ into the quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $
$ x=\frac{-(-2)\pm\sqrt{24}}{2\times1} $
Simplify $ \sqrt{24}=2\sqrt{6} $, so:
$ x=\frac{2\pm2\sqrt{6}}{2} $
$ x = 1\pm\sqrt{6} $

# Answer:
The zeros of the quadratic equation $ x^2-2x - 5=0 $ are $ x = 1+\sqrt{6}\approx1 + 2.45=3.45 $ and $ x=1-\sqrt{6}\approx1-2.45=-1.45 $ (approximate values)

Let's solve the third quadratic equation $ x^2-4x - 98=0 $ (derived from $ x^2-4x - 91=7\Rightarrow x^2-4x - 98 = 0 $) using the quadratic formula.

# Explanation:
## Step 1: Identify $ a $, $ b $, $ c $
For the quadratic equation $ ax^2+bx + c = 0 $, in $ x^2-4x - 98=0 $, we have $ a = 1 $, $ b=-4 $, $ c=-98 $

## Step 2: Calculate the discriminant $ D=b^2-4ac $
Substitute $ a = 1 $, $ b=-4 $, $ c=-98 $ into the discriminant formula:
$ D=(-4)^2-4\times1\times(-98) $
$ D = 16 + 392 $
$ D = 408 $

## Step 3: Apply the quadratic formula
Substitute $ a = 1 $, $ b=-4 $, $ D = 408 $ into the quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $
$ x=\frac{-(-4)\pm\sqrt{408}}{2\times1} $
Simplify $ \sqrt{408}=\sqrt{4\times102}=2\sqrt{102} $, so:
$ x=\frac{4\pm2\sqrt{102}}{2} $
$ x = 2\pm\sqrt{102} $

# Answer:
The zeros of the quadratic equation $ x^2-4x - 98=0 $ are $ x=2+\sqrt{102}\approx2 + 10.1=12.1 $ and $ x=2-\sqrt{102}\approx2-10.1=-8.1 $ (approximate values)

We can solve the fourth quadratic equation $ x^2-10x + 18=0 $ (derived from $ x^2-10x + 26 = 8\Rightarrow x^2-10x + 18=0 $) using the quadratic formula.

# Explanation:
## Step 1: Identify $ a $, $ b $, $ c $
For the quadratic equation $ ax^2+bx + c = 0 $, in $ x^2-10x + 18=0 $, we have $ a = 1 $, $ b=-10 $, $ c = 18 $

## Step 2: Calculate the discriminant $ D=b^2-4ac $
Substitute $ a = 1 $, $ b=-10 $, $ c = 18 $ into the discriminant formula:
$ D=(-10)^2-4\times1\times18 $
$ D = 100 - 72 $
$ D = 28 $

## Step 3: Apply the quadratic formula
Substitute $ a = 1 $, $ b=-10 $, $ D = 28 $ into the quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $
$ x=\frac{-(-10)\pm\sqrt{28}}{2\times1} $
Simplify $ \sqrt{28}=\sqrt{4\times7}=2\sqrt{7} $, so:
$ x=\frac{10\pm2\sqrt{7}}{2} $
$ x = 5\pm\sqrt{7} $

# Answer:
The zeros of the quadratic equation $ x^2-10x + 18=0 $ are $ x=5+\sqrt{7}\approx5 + 2.65=7.65 $ and $ x=5-\sqrt{7}\approx5-2.65=2.35 $ (approximate values)

We can solve the fifth quadratic equation $ x^2-4x - 4=0 $ (derived from $ x^2-4x + 1=5\Rightarrow x^2-4x - 4=0 $) using the quadratic formula.

# Explanation:
## Step 1: Identify $ a $, $ b $, $ c $
For the quadratic equation $ ax^2+bx + c = 0 $, in $ x^2-4x - 4=0 $, we have $ a = 1 $, $ b=-4 $, $ c=-4 $

## Step 2: Calculate the discriminant $ D=b^2-4ac $
Substitute $ a = 1 $, $ b=-4 $, $ c=-4 $ into the discriminant formula:
$ D=(-4)^2-4\times1\times(-4) $
$ D = 16 + 16 $
$ D = 32 $

## Step 3: Apply the quadratic formula
Substitute $ a = 1 $, $ b=-4 $, $ D = 32 $ into the quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $
$ x=\frac{-(-4)\pm\sqrt{32}}{2\times1} $
Simplify $ \sqrt{32}=\sqrt{16\times2}=4\sqrt{2} $, so:
$ x=\frac{4\pm4\sqrt{2}}{2} $
$ x = 2\pm2\sqrt{2} $

# Answer:
The zeros of the quadratic equation $ x^2-4x - 4=0 $ are $ x=2 + 2\sqrt{2}\approx2+2.83 = 4.83 $ and $ x=2-2\sqrt{2}\approx2 - 2.83=-0.83 $ (approximate values)

We can solve the sixth quadratic equation $ x^2-2x + 20=0 $ (derived from $ x^2-2x=-20\Rightarrow x^2-2x + 20=0 $) using the quadratic formula.

# Explanation:
## Step 1: Identify $ a $, $ b $, $ c $
For the quadratic equation $ ax^2+bx + c = 0 $, in $ x^2-2x + 20=0 $, we have $ a = 1 $, $ b=-2 $, $ c = 20 $

## Step 2: Calculate the discriminant $ D=b^2-4ac $
Substitute $ a = 1 $, $ b=-2 $, $ c = 20 $ into the discriminant formula:
$ D=(-2)^2-4\times1\times20 $
$ D = 4 - 80 $
$ D=-76 $

Since the discriminant $ D=-76<0 $, the roots are complex numbers. Using the fact that $ \sqrt{-76}=\sqrt{76}i=\sqrt{4\times19}i = 2\sqrt{19}i $

## Step 3: Apply the quadratic formula
Substitute $ a = 1 $, $ b=-2 $, $ D=-76 $ into the quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $
$ x=\frac{-(-2)\pm\sqrt{-76}}{2\times1} $
$ x=\frac{2\pm2\sqrt{19}i}{2} $
$ x = 1\pm\sqrt{19}i $

# Answer:
The zeros of the quadratic equation $ x^2-2x + 20=0 $ are $ x = 1+\sqrt{19}i $ and $ x=1-\sqrt{19}i $ (complex roots)

We can solve the seventh quadratic equation $ x^2-18x - 40=0 $ (derived from $ x^2=18x + 40\Rightarrow x^2-18x - 40=0 $) using the quadratic formula.

# Explanation:
## Step 1: Identify $ a $, $ b $, $ c $
For the quadratic equation $ ax^2+bx + c = 0 $, in $ x^2-18x - 40=0 $, we have $ a = 1 $, $ b=-18 $, $ c=-40 $

## Step 2: Calculate the discriminant $ D=b^2-4ac $
Substitute $ a = 1 $, $ b=-18 $, $ c=-40 $ into the discriminant formula:
$ D=(-18)^2-4\times1\times(-40) $
$ D = 324+160 $
$ D = 484 $

## Step 3: Apply the quadratic formula
Substitute $ a = 1 $, $ b=-18 $, $ D = 484 $ into the quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $
$ x=\frac{-(-18)\pm\sqrt{484}}{2\times1} $
Since $ \sqrt{484} = 22 $, we have:
$ x=\frac{18\pm22}{2} $

## Step 4: Find the two solutions
For the plus sign:
$ x=\frac{18 + 22}{2}=\frac{40}{2}=20 $
For the minus sign:
$ x=\frac{18-22}{2}=\frac{-4}{2}=-2 $

# Answer:
The zeros of the quadratic equation $ x^2-18x - 40=0 $ are $ x = 20 $ and $ x=-2 $

We can solve the eighth quadratic equation $ 8x^2+16x - 42=0 $ or simplified as $ 4x^2+8x - 21=0 $ (dividing both sides by 2) using the quadratic formula.

# Explanation:
## Step 1: Identify $ a $, $ b $, $ c $
For the quadratic equation $ ax^2+bx + c = 0 $, in $ 4x^2+8x - 21=0 $, we have $ a = 4 $, $ b = 8 $, $ c=-21 $

## Step 2: Calculate the discriminant $ D=b^2-4ac $
Substitute $ a = 4 $, $ b = 8 $, $ c=-21 $ into the discriminant formula:
$ D=(8)^2-4\times4\times(-21) $
$ D = 64 + 336 $
$ D = 400 $

## Step 3: Apply the quadratic formula
Substitute $ a = 4 $, $ b = 8 $, $ D = 400 $ into the quadratic formula $ x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $
$ x=\frac{-8\pm\sqrt{400}}{2\times4} $
Since $ \sqrt{400}=20 $, we have:
$ x=\frac{-8\pm20}{8} $

## Step 4: Find the two solutions
For the plus sign:
$ x=\frac{-8 + 20}{8}=\frac{12}{8}=\frac{3}{2}=1.5 $
For the minus sign:
$ x=\frac{-8-20}{8}=\frac{-28}{8}=-\frac{7}{2}=-3.5 $

# Answer:
The zeros of the quadratic equation $ 8x^2+16x - 42=0 $ are $ x=\frac{3}{2} $ and $ x=-\frac{7}{2} $

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QUESTION IMAGE

Question

Explanation:

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

Step 2: Calculate the discriminant \( D = b^2 - 4ac \)

Step 3: Apply the quadratic formula

Step 4: Find the two solutions

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

Step 2: Calculate the discriminant \( D=b^2-4ac \)

Step 3: Apply the quadratic formula

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

Step 2: Calculate the discriminant \( D=b^2-4ac \)

Step 3: Apply the quadratic formula

Answer: