Question
Using the quadratic formula, find the solution to x2+6x+9=0
.(1 point)
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.(1 point)
Remaining Attempts : 2View previous attempt
Answers
Solve −2x2+3x−1=−2
using the quadratic formula. Round the solution(s) to the nearest hundredths. Enter the smaller solution first.(1 point)
using the quadratic formula. Round the solution(s) to the nearest hundredths. Enter the smaller solution first.(1 point)
GPT-4o mini
To use the quadratic formula, we first need to express both equations in standard form \( ax^2 + bx + c = 0 \).
### First Equation: \(x^2 + 6x + 9 = 0\)
This equation is already in standard form with:
- \( a = 1 \)
- \( b = 6 \)
- \( c = 9 \)
Now, we apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Calculating the discriminant (\(b^2 - 4ac\)):
\[
b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0
\]
Since the discriminant is 0, there is one real solution:
\[
x = \frac{-6 \pm \sqrt{0}}{2 \cdot 1} = \frac{-6}{2} = -3
\]
### Second Equation: \(-2x^2 + 3x - 1 = -2\)
First, we need to rearrange the equation to standard form:
\[
-2x^2 + 3x - 1 + 2 = 0 \implies -2x^2 + 3x + 1 = 0
\]
Now, multiply the equation by \(-1\) to make \(a\) positive:
\[
2x^2 - 3x - 1 = 0
\]
In this case, we have:
- \( a = 2 \)
- \( b = -3 \)
- \( c = -1 \)
Now, we apply the quadratic formula again:
\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}
\]
Calculating the discriminant:
\[
(-3)^2 - 4 \cdot 2 \cdot (-1) = 9 + 8 = 17
\]
Now, substituting back into the formula:
\[
x = \frac{3 \pm \sqrt{17}}{4}
\]
Calculating the two solutions:
1. The smaller solution:
\[
x_1 = \frac{3 - \sqrt{17}}{4}
\]
2. The larger solution:
\[
x_2 = \frac{3 + \sqrt{17}}{4}
\]
Now, we round the solutions to the nearest hundredths:
- First, calculate \( \sqrt{17} \approx 4.123 \)
- Then,
\[
x_1 = \frac{3 - 4.123}{4} \approx \frac{-1.123}{4} \approx -0.281
\]
Rounding: \( -0.28 \)
\[
x_2 = \frac{3 + 4.123}{4} \approx \frac{7.123}{4} \approx 1.781
\]
Rounding: \( 1.78 \)
### Final Answers:
1. The solution to \( x^2 + 6x + 9 = 0 \) is \( x = -3 \).
2. The solutions to \(-2x^2 + 3x - 1 = -2\) are approximately \( x_1 \approx -0.28 \) and \( x_2 \approx 1.78 \). Entering the smaller solution first:
- \( -0.28, 1.78 \)
### First Equation: \(x^2 + 6x + 9 = 0\)
This equation is already in standard form with:
- \( a = 1 \)
- \( b = 6 \)
- \( c = 9 \)
Now, we apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Calculating the discriminant (\(b^2 - 4ac\)):
\[
b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0
\]
Since the discriminant is 0, there is one real solution:
\[
x = \frac{-6 \pm \sqrt{0}}{2 \cdot 1} = \frac{-6}{2} = -3
\]
### Second Equation: \(-2x^2 + 3x - 1 = -2\)
First, we need to rearrange the equation to standard form:
\[
-2x^2 + 3x - 1 + 2 = 0 \implies -2x^2 + 3x + 1 = 0
\]
Now, multiply the equation by \(-1\) to make \(a\) positive:
\[
2x^2 - 3x - 1 = 0
\]
In this case, we have:
- \( a = 2 \)
- \( b = -3 \)
- \( c = -1 \)
Now, we apply the quadratic formula again:
\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}
\]
Calculating the discriminant:
\[
(-3)^2 - 4 \cdot 2 \cdot (-1) = 9 + 8 = 17
\]
Now, substituting back into the formula:
\[
x = \frac{3 \pm \sqrt{17}}{4}
\]
Calculating the two solutions:
1. The smaller solution:
\[
x_1 = \frac{3 - \sqrt{17}}{4}
\]
2. The larger solution:
\[
x_2 = \frac{3 + \sqrt{17}}{4}
\]
Now, we round the solutions to the nearest hundredths:
- First, calculate \( \sqrt{17} \approx 4.123 \)
- Then,
\[
x_1 = \frac{3 - 4.123}{4} \approx \frac{-1.123}{4} \approx -0.281
\]
Rounding: \( -0.28 \)
\[
x_2 = \frac{3 + 4.123}{4} \approx \frac{7.123}{4} \approx 1.781
\]
Rounding: \( 1.78 \)
### Final Answers:
1. The solution to \( x^2 + 6x + 9 = 0 \) is \( x = -3 \).
2. The solutions to \(-2x^2 + 3x - 1 = -2\) are approximately \( x_1 \approx -0.28 \) and \( x_2 \approx 1.78 \). Entering the smaller solution first:
- \( -0.28, 1.78 \)