Answer:
(- 1, 2) and (- [tex]\frac{1}{2}[/tex], [tex]\frac{1}{2}[/tex])
Step-by-step explanation:
Given the 2 equations
y = 2x² → (1)
y = - 3x - 1 → (2)
Substitute y = 2x² into (2)
2x² = - 3x - 1 ( add 3x and 1 to both sides )
2x² + 3x + 1 = 0 ← in standard form
(2x + 1)(x + 1) = 0 ← in factored form
Equate each factor to zero and solve for x
2x + 1 = 0 ⇒ 2x = - 1 ⇒ x = - [tex]\frac{1}{2}[/tex]
x + 1 = 0 ⇒ x = - 1
Substitute these values into (2) for corresponding values of y
x = - 1 → y = - 3(- 1) - 1 = 3 - 1 = 2 ⇒ (- 1, 2 )
x = - [tex]\frac{1}{2}[/tex] → y = [tex]\frac{3}{2}[/tex] - 1 = [tex]\frac{1}{2}[/tex] ⇒ (- [tex]\frac{1}{2}[/tex], [tex]\frac{1}{2}[/tex] )
We are given a system of equations.
Doing this, we get that the solutions of the system are: [tex](-\frac{1}{2},\frac{1}{2}), (-1,2)[/tex]
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
y = 2x³
y = –3x −1
Equaling them, we get:
[tex]2x^2 = -3x - 1[/tex]
[tex]2x^2 + 3x + 1 = 0[/tex]
Which is a quadratic equation with [tex]a = 2, b = 3, c = 1[/tex]
We solve it to find the values of x.
[tex]\Delta = 3^{2} - 4(2)(1) = 9 - 8 = 1[/tex]
[tex]x_{1} = \frac{-3 + \sqrt{1}}{2*(2)} = -\frac{2}{4} = -\frac{1}{2}[/tex]
[tex]x_{2} = \frac{-3 - \sqrt{1}}{2*2} = -1[/tex]
Solutions of y:
For x = -1/2:
[tex]y = 2(-\frac{1}{2})^2 = 2(\frac{1}{4}) = \frac{1}{2}[/tex]
Thus, one of the solutions is: [tex](-\frac{1}{2},\frac{1}{2})[/tex]
For x = -1:
[tex]y = 2(-1)^2 = 2[/tex]
Thus, the other solution is: (-1,2).
Then, the two solutions are: [tex](-\frac{1}{2},\frac{1}{2}), (-1,2)[/tex]
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