Asked by T-Swizzle
Use the substitution method to find the solutions to the linear-quadratic system: y = x^2 - x - 40, y = -3x - 5
Answers
Answered by
GPT-4o mini
To solve the linear-quadratic system of equations using the substitution method, we have the following equations:
1. \( y = x^2 - x - 40 \) (quadratic equation)
2. \( y = -3x - 5 \) (linear equation)
We can substitute the expression for \( y \) from the linear equation into the quadratic equation.
Setting the two equations equal to each other, we have:
\[
-3x - 5 = x^2 - x - 40
\]
Next, let's rearrange this equation:
1. Move all terms to one side of the equation:
\[
x^2 - x + 3x + 40 - 5 = 0
\]
2. Combine like terms:
\[
x^2 + 2x + 35 = 0
\]
Now we will solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a = 1\), \(b = 2\), and \(c = 35\). First, we need to calculate the discriminant (\(b^2 - 4ac\)):
\[
b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 35 = 4 - 140 = -136
\]
Since the discriminant is negative, this tells us that there are no real solutions to the quadratic equation \(x^2 + 2x + 35 = 0\). Therefore, the original system of equations has no real solutions.
In conclusion, the linear-quadratic system \(y = x^2 - x - 40\) and \(y = -3x - 5\) does not intersect in the real number plane, meaning there are no solutions to this system.
1. \( y = x^2 - x - 40 \) (quadratic equation)
2. \( y = -3x - 5 \) (linear equation)
We can substitute the expression for \( y \) from the linear equation into the quadratic equation.
Setting the two equations equal to each other, we have:
\[
-3x - 5 = x^2 - x - 40
\]
Next, let's rearrange this equation:
1. Move all terms to one side of the equation:
\[
x^2 - x + 3x + 40 - 5 = 0
\]
2. Combine like terms:
\[
x^2 + 2x + 35 = 0
\]
Now we will solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a = 1\), \(b = 2\), and \(c = 35\). First, we need to calculate the discriminant (\(b^2 - 4ac\)):
\[
b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 35 = 4 - 140 = -136
\]
Since the discriminant is negative, this tells us that there are no real solutions to the quadratic equation \(x^2 + 2x + 35 = 0\). Therefore, the original system of equations has no real solutions.
In conclusion, the linear-quadratic system \(y = x^2 - x - 40\) and \(y = -3x - 5\) does not intersect in the real number plane, meaning there are no solutions to this system.