Answer :
To find the solutions for the quadratic equation \( 4x^2 + 16x + 7 = 0 \), we will use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients of the quadratic equation are:
[tex]\[ a = 4,\quad b = 16,\quad c = 7 \][/tex]
### Step-by-Step Solution
1. Calculate the Discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of \(a\), \(b\), and \(c\):
[tex]\[ \Delta = 16^2 - 4 \cdot 4 \cdot 7 = 256 - 112 = 144 \][/tex]
2. Calculate the Two Possible Values of \(x\):
[tex]\[ x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
- Calculate \( x_1 \):
[tex]\[ x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-16 + \sqrt{144}}{2 \cdot 4} = \frac{-16 + 12}{8} = \frac{-4}{8} = -0.5 \][/tex]
- Calculate \( x_2 \):
[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-16 - \sqrt{144}}{2 \cdot 4} = \frac{-16 - 12}{8} = \frac{-28}{8} = -3.5 \][/tex]
3. Round the Solutions:
The solutions are already in the form requiring rounding to 1 decimal place:
[tex]\[ x_1 = -0.5 \quad \text{and} \quad x_2 = -3.5 \][/tex]
### Final Answer
The two possible values of \( x \) for the quadratic equation \( 4x^2 + 16x + 7 = 0 \) rounded to 1 decimal place are:
[tex]\[ x_1 = -0.5 \quad \text{and} \quad x_2 = -3.5 \][/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients of the quadratic equation are:
[tex]\[ a = 4,\quad b = 16,\quad c = 7 \][/tex]
### Step-by-Step Solution
1. Calculate the Discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of \(a\), \(b\), and \(c\):
[tex]\[ \Delta = 16^2 - 4 \cdot 4 \cdot 7 = 256 - 112 = 144 \][/tex]
2. Calculate the Two Possible Values of \(x\):
[tex]\[ x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
- Calculate \( x_1 \):
[tex]\[ x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-16 + \sqrt{144}}{2 \cdot 4} = \frac{-16 + 12}{8} = \frac{-4}{8} = -0.5 \][/tex]
- Calculate \( x_2 \):
[tex]\[ x_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-16 - \sqrt{144}}{2 \cdot 4} = \frac{-16 - 12}{8} = \frac{-28}{8} = -3.5 \][/tex]
3. Round the Solutions:
The solutions are already in the form requiring rounding to 1 decimal place:
[tex]\[ x_1 = -0.5 \quad \text{and} \quad x_2 = -3.5 \][/tex]
### Final Answer
The two possible values of \( x \) for the quadratic equation \( 4x^2 + 16x + 7 = 0 \) rounded to 1 decimal place are:
[tex]\[ x_1 = -0.5 \quad \text{and} \quad x_2 = -3.5 \][/tex]
