Question
Solve: \( 4 x^{2}+8 x+1=0 \)
Ask by Sullivan Lambert. in the United States
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The solutions are \( x = -1 + \frac{\sqrt{3}}{2} \) and \( x = -1 - \frac{\sqrt{3}}{2} \).
Solution
To solve the quadratic equation:
\[ 4x^{2} + 8x + 1 = 0 \]
we can use the **quadratic formula**, which is given by:
\[
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
\]
where \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^{2} + bx + c = 0 \).
### Step 1: Identify the coefficients
For the equation \( 4x^{2} + 8x + 1 = 0 \):
- \( a = 4 \)
- \( b = 8 \)
- \( c = 1 \)
### Step 2: Compute the discriminant
The discriminant (\( \Delta \)) is the part under the square root in the quadratic formula:
\[
\Delta = b^{2} - 4ac
\]
Plugging in the values:
\[
\Delta = 8^{2} - 4(4)(1) = 64 - 16 = 48
\]
### Step 3: Apply the quadratic formula
Now, substitute \( a \), \( b \), and \( \Delta \) into the quadratic formula:
\[
x = \frac{-8 \pm \sqrt{48}}{2 \times 4} = \frac{-8 \pm \sqrt{16 \times 3}}{8} = \frac{-8 \pm 4\sqrt{3}}{8}
\]
### Step 4: Simplify the expression
Divide numerator and denominator by 4:
\[
x = \frac{-2 \pm \sqrt{3}}{2}
\]
Alternatively, this can be written as:
\[
x = -1 \pm \frac{\sqrt{3}}{2}
\]
### Final Answer
The solutions to the equation \( 4x^{2} + 8x + 1 = 0 \) are:
\[
x = -1 + \frac{\sqrt{3}}{2} \quad \text{and} \quad x = -1 - \frac{\sqrt{3}}{2}
\]
Or, equivalently:
\[
x = \frac{-2 \pm \sqrt{3}}{2}
\]
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Bonus Knowledge
To solve the quadratic equation \( 4x^2 + 8x + 1 = 0 \), we can use the quadratic formula, which states \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 4 \), \( b = 8 \), and \( c = 1 \). First, calculate the discriminant: \[ b^2 - 4ac = 8^2 - 4 \cdot 4 \cdot 1 = 64 - 16 = 48 \] Now, plug the values into the quadratic formula: \[ x = \frac{-8 \pm \sqrt{48}}{2 \cdot 4} = \frac{-8 \pm 4\sqrt{3}}{8} = \frac{-2 \pm \sqrt{3}}{2} \] Thus, the solutions are: \[ x = -1 + \frac{\sqrt{3}}{2} \quad \text{or} \quad x = -1 - \frac{\sqrt{3}}{2} \]