Question
Solve the following quadratic equations using completing the square
1. x² - 4x
2. x² + 8x - 5 = 0
3. 4x² + 32x + 68 =0
4. y = x² - 8x + 18 in the vertex form
1. x² - 4x
2. x² + 8x - 5 = 0
3. 4x² + 32x + 68 =0
4. y = x² - 8x + 18 in the vertex form
Answers
Answered by
GPT-4o mini
To solve the provided quadratic equations using the completing the square method, we will follow the standard steps for each equation.
### 1. \( x^2 - 4x = 0 \)
Step 1: Move to one side of the equation:
\[
x^2 - 4x = 0
\]
Step 2: Complete the square:
\[
x^2 - 4x + 4 = 4
\]
Now we can rewrite it as:
\[
(x - 2)^2 = 4
\]
Step 3: Solve for \(x\):
\[
x - 2 = \pm 2
\]
Thus,
\[
x - 2 = 2 \implies x = 4 \quad \text{and} \quad x - 2 = -2 \implies x = 0
\]
Final solutions:
\[
x = 4 \quad \text{and} \quad x = 0
\]
---
### 2. \( x^2 + 8x - 5 = 0 \)
Step 1: Move the constant to the other side:
\[
x^2 + 8x = 5
\]
Step 2: Complete the square:
Take half of 8, square it:
\(\left(\frac{8}{2}\right)^2 = 16\).
Add 16 to both sides:
\[
x^2 + 8x + 16 = 5 + 16
\]
This simplifies to:
\[
(x + 4)^2 = 21
\]
Step 3: Solve for \(x\):
\[
x + 4 = \pm \sqrt{21}
\]
Thus:
\[
x = -4 + \sqrt{21} \quad \text{and} \quad x = -4 - \sqrt{21}
\]
Final solutions:
\[
x = -4 + \sqrt{21} \quad \text{and} \quad x = -4 - \sqrt{21}
\]
---
### 3. \( 4x^2 + 32x + 68 = 0 \)
Step 1: Divide the whole equation by 4 to simplify:
\[
x^2 + 8x + 17 = 0
\]
Step 2: Move the constant to the other side:
\[
x^2 + 8x = -17
\]
Step 3: Complete the square:
Take half of 8, square it:
\(\left(\frac{8}{2}\right)^2 = 16\).
Add 16 to both sides:
\[
x^2 + 8x + 16 = -17 + 16
\]
This simplifies to:
\[
(x + 4)^2 = -1
\]
Step 4: Solve for \(x\):
\[
x + 4 = \pm i
\]
Thus:
\[
x = -4 + i \quad \text{and} \quad x = -4 - i
\]
Final solutions:
\[
x = -4 + i \quad \text{and} \quad x = -4 - i
\]
---
### 4. \( y = x^2 - 8x + 18 \)
To write this in vertex form \(y = a(x - h)^2 + k\):
Step 1: Complete the square for the quadratic expression:
Take half of -8, square it:
\(\left(\frac{-8}{2}\right)^2 = 16\).
Add and subtract 16:
\[
y = x^2 - 8x + 16 - 16 + 18
\]
This simplifies to:
\[
y = (x - 4)^2 + 2
\]
Now, the vertex form is:
\[
y = (x - 4)^2 + 2
\]
Final answer:
\[
y = (x - 4)^2 + 2
\]
This indicates the vertex of the parabola is at \((4, 2)\).
### 1. \( x^2 - 4x = 0 \)
Step 1: Move to one side of the equation:
\[
x^2 - 4x = 0
\]
Step 2: Complete the square:
\[
x^2 - 4x + 4 = 4
\]
Now we can rewrite it as:
\[
(x - 2)^2 = 4
\]
Step 3: Solve for \(x\):
\[
x - 2 = \pm 2
\]
Thus,
\[
x - 2 = 2 \implies x = 4 \quad \text{and} \quad x - 2 = -2 \implies x = 0
\]
Final solutions:
\[
x = 4 \quad \text{and} \quad x = 0
\]
---
### 2. \( x^2 + 8x - 5 = 0 \)
Step 1: Move the constant to the other side:
\[
x^2 + 8x = 5
\]
Step 2: Complete the square:
Take half of 8, square it:
\(\left(\frac{8}{2}\right)^2 = 16\).
Add 16 to both sides:
\[
x^2 + 8x + 16 = 5 + 16
\]
This simplifies to:
\[
(x + 4)^2 = 21
\]
Step 3: Solve for \(x\):
\[
x + 4 = \pm \sqrt{21}
\]
Thus:
\[
x = -4 + \sqrt{21} \quad \text{and} \quad x = -4 - \sqrt{21}
\]
Final solutions:
\[
x = -4 + \sqrt{21} \quad \text{and} \quad x = -4 - \sqrt{21}
\]
---
### 3. \( 4x^2 + 32x + 68 = 0 \)
Step 1: Divide the whole equation by 4 to simplify:
\[
x^2 + 8x + 17 = 0
\]
Step 2: Move the constant to the other side:
\[
x^2 + 8x = -17
\]
Step 3: Complete the square:
Take half of 8, square it:
\(\left(\frac{8}{2}\right)^2 = 16\).
Add 16 to both sides:
\[
x^2 + 8x + 16 = -17 + 16
\]
This simplifies to:
\[
(x + 4)^2 = -1
\]
Step 4: Solve for \(x\):
\[
x + 4 = \pm i
\]
Thus:
\[
x = -4 + i \quad \text{and} \quad x = -4 - i
\]
Final solutions:
\[
x = -4 + i \quad \text{and} \quad x = -4 - i
\]
---
### 4. \( y = x^2 - 8x + 18 \)
To write this in vertex form \(y = a(x - h)^2 + k\):
Step 1: Complete the square for the quadratic expression:
Take half of -8, square it:
\(\left(\frac{-8}{2}\right)^2 = 16\).
Add and subtract 16:
\[
y = x^2 - 8x + 16 - 16 + 18
\]
This simplifies to:
\[
y = (x - 4)^2 + 2
\]
Now, the vertex form is:
\[
y = (x - 4)^2 + 2
\]
Final answer:
\[
y = (x - 4)^2 + 2
\]
This indicates the vertex of the parabola is at \((4, 2)\).
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