Answer :
To solve the system of equations given:
1. Write down the equations:
- Equation 1: [tex]\( y = 20x^2 + 6 \)[/tex]
- Equation 2: [tex]\( y = -23x \)[/tex]
2. Set the equations equal to each other:
Since both expressions on the right-hand side equal [tex]\( y \)[/tex], you can start by setting them equal to each other:
[tex]\[
20x^2 + 6 = -23x
\][/tex]
3. Rearrange the equation:
Move all the terms to one side to form a quadratic equation:
[tex]\[
20x^2 + 23x + 6 = 0
\][/tex]
4. Solve the quadratic equation:
To solve the quadratic equation, you can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 20 \)[/tex], [tex]\( b = 23 \)[/tex], and [tex]\( c = 6 \)[/tex].
5. Calculate the discriminant:
First, calculate [tex]\( b^2 - 4ac \)[/tex]:
[tex]\[
b^2 - 4ac = 23^2 - 4 \times 20 \times 6 = 529 - 480 = 49
\][/tex]
Since the discriminant is positive, there are two real solutions.
6. Find the roots using the quadratic formula:
[tex]\[
x = \frac{-23 \pm \sqrt{49}}{40}
\][/tex]
This simplifies to:
[tex]\[
x = \frac{-23 \pm 7}{40}
\][/tex]
- For the positive root:
[tex]\[
x = \frac{-23 + 7}{40} = \frac{-16}{40} = -\frac{2}{5}
\][/tex]
- For the negative root:
[tex]\[
x = \frac{-23 - 7}{40} = \frac{-30}{40} = -\frac{3}{4}
\][/tex]
7. Find the corresponding [tex]\( y \)[/tex] values:
Substitute each [tex]\( x \)[/tex] value back into equation 2 ([tex]\( y = -23x \)[/tex]) to find [tex]\( y \)[/tex].
- For [tex]\( x = -\frac{2}{5} \)[/tex]:
[tex]\[
y = -23 \left(-\frac{2}{5}\right) = \frac{46}{5}
\][/tex]
- For [tex]\( x = -\frac{3}{4} \)[/tex]:
[tex]\[
y = -23 \left(-\frac{3}{4}\right) = \frac{69}{4}
\][/tex]
8. State the solutions:
The solutions to the system, based on the calculations, are:
- [tex]\(\left(-\frac{3}{4}, \frac{69}{4}\right)\)[/tex]
- [tex]\(\left(-\frac{2}{5}, \frac{46}{5}\right)\)[/tex]
So the correct answer is B: [tex]\(\left(-\frac{3}{4}, \frac{69}{4}\right)\)[/tex] and [tex]\(\left(\frac{2}{5}, \frac{46}{5}\right)\)[/tex].
1. Write down the equations:
- Equation 1: [tex]\( y = 20x^2 + 6 \)[/tex]
- Equation 2: [tex]\( y = -23x \)[/tex]
2. Set the equations equal to each other:
Since both expressions on the right-hand side equal [tex]\( y \)[/tex], you can start by setting them equal to each other:
[tex]\[
20x^2 + 6 = -23x
\][/tex]
3. Rearrange the equation:
Move all the terms to one side to form a quadratic equation:
[tex]\[
20x^2 + 23x + 6 = 0
\][/tex]
4. Solve the quadratic equation:
To solve the quadratic equation, you can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 20 \)[/tex], [tex]\( b = 23 \)[/tex], and [tex]\( c = 6 \)[/tex].
5. Calculate the discriminant:
First, calculate [tex]\( b^2 - 4ac \)[/tex]:
[tex]\[
b^2 - 4ac = 23^2 - 4 \times 20 \times 6 = 529 - 480 = 49
\][/tex]
Since the discriminant is positive, there are two real solutions.
6. Find the roots using the quadratic formula:
[tex]\[
x = \frac{-23 \pm \sqrt{49}}{40}
\][/tex]
This simplifies to:
[tex]\[
x = \frac{-23 \pm 7}{40}
\][/tex]
- For the positive root:
[tex]\[
x = \frac{-23 + 7}{40} = \frac{-16}{40} = -\frac{2}{5}
\][/tex]
- For the negative root:
[tex]\[
x = \frac{-23 - 7}{40} = \frac{-30}{40} = -\frac{3}{4}
\][/tex]
7. Find the corresponding [tex]\( y \)[/tex] values:
Substitute each [tex]\( x \)[/tex] value back into equation 2 ([tex]\( y = -23x \)[/tex]) to find [tex]\( y \)[/tex].
- For [tex]\( x = -\frac{2}{5} \)[/tex]:
[tex]\[
y = -23 \left(-\frac{2}{5}\right) = \frac{46}{5}
\][/tex]
- For [tex]\( x = -\frac{3}{4} \)[/tex]:
[tex]\[
y = -23 \left(-\frac{3}{4}\right) = \frac{69}{4}
\][/tex]
8. State the solutions:
The solutions to the system, based on the calculations, are:
- [tex]\(\left(-\frac{3}{4}, \frac{69}{4}\right)\)[/tex]
- [tex]\(\left(-\frac{2}{5}, \frac{46}{5}\right)\)[/tex]
So the correct answer is B: [tex]\(\left(-\frac{3}{4}, \frac{69}{4}\right)\)[/tex] and [tex]\(\left(\frac{2}{5}, \frac{46}{5}\right)\)[/tex].
