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To determine which of the given choices are solutions to the quadratic equation [tex]\(2x^2 + 6x - 10 = x^2 + 6\)[/tex], let's follow a systematic approach to solve it.
First, we need to simplify and rearrange the equation:
### Step 1: Simplify the Equation
Original equation:
[tex]\[ 2x^2 + 6x - 10 = x^2 + 6 \][/tex]
Subtract [tex]\(x^2 + 6\)[/tex] from both sides to set the equation to zero:
[tex]\[ 2x^2 + 6x - 10 - x^2 - 6 = 0 \][/tex]
Combine like terms:
[tex]\[ 2x^2 - x^2 + 6x - 10 - 6 = 0 \][/tex]
[tex]\[ x^2 + 6x - 16 = 0 \][/tex]
Now we have the simplified quadratic equation:
[tex]\[ x^2 + 6x - 16 = 0 \][/tex]
### Step 2: Solve the Quadratic Equation
Quadratic equations of the form [tex]\(ax^2 + bx + c = 0\)[/tex] can be solved using factoring, completing the square, or the quadratic formula. In this instance, we will solve it to find the roots (solutions).
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\(x^2 + 6x - 16 = 0\)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 6\)[/tex]
- [tex]\(c = -16\)[/tex]
Plugging these values into the quadratic formula:
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 + 64}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm 10}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-6 + 10}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ x = \frac{-6 - 10}{2} = \frac{-16}{2} = -8 \][/tex]
So the solutions to the equation [tex]\(x^2 + 6x - 16 = 0\)[/tex] are:
[tex]\[ x = -8 \text{ and } x = 2 \][/tex]
### Step 3: Verify the Solutions with Given Choices
Given choices are:
- A. -8
- B. -2
- C. [tex]\(\frac{1}{3}\)[/tex]
- D. 8
- E. 2
- F. [tex]\(-\frac{1}{2}\)[/tex]
We see that the solutions [tex]\(-8\)[/tex] and [tex]\(2\)[/tex] are among the choices.
### Conclusion
The choices that are solutions to the quadratic equation [tex]\(2x^2 + 6x - 10 = x^2 + 6\)[/tex] are:
- A. -8
- E. 2
First, we need to simplify and rearrange the equation:
### Step 1: Simplify the Equation
Original equation:
[tex]\[ 2x^2 + 6x - 10 = x^2 + 6 \][/tex]
Subtract [tex]\(x^2 + 6\)[/tex] from both sides to set the equation to zero:
[tex]\[ 2x^2 + 6x - 10 - x^2 - 6 = 0 \][/tex]
Combine like terms:
[tex]\[ 2x^2 - x^2 + 6x - 10 - 6 = 0 \][/tex]
[tex]\[ x^2 + 6x - 16 = 0 \][/tex]
Now we have the simplified quadratic equation:
[tex]\[ x^2 + 6x - 16 = 0 \][/tex]
### Step 2: Solve the Quadratic Equation
Quadratic equations of the form [tex]\(ax^2 + bx + c = 0\)[/tex] can be solved using factoring, completing the square, or the quadratic formula. In this instance, we will solve it to find the roots (solutions).
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\(x^2 + 6x - 16 = 0\)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 6\)[/tex]
- [tex]\(c = -16\)[/tex]
Plugging these values into the quadratic formula:
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 + 64}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm 10}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-6 + 10}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ x = \frac{-6 - 10}{2} = \frac{-16}{2} = -8 \][/tex]
So the solutions to the equation [tex]\(x^2 + 6x - 16 = 0\)[/tex] are:
[tex]\[ x = -8 \text{ and } x = 2 \][/tex]
### Step 3: Verify the Solutions with Given Choices
Given choices are:
- A. -8
- B. -2
- C. [tex]\(\frac{1}{3}\)[/tex]
- D. 8
- E. 2
- F. [tex]\(-\frac{1}{2}\)[/tex]
We see that the solutions [tex]\(-8\)[/tex] and [tex]\(2\)[/tex] are among the choices.
### Conclusion
The choices that are solutions to the quadratic equation [tex]\(2x^2 + 6x - 10 = x^2 + 6\)[/tex] are:
- A. -8
- E. 2
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