is (0,0) a solution to this system y> x2+x-4 y <x2+2x+1

Answer:
Yes, the ordered pair (0,0) is a solution of the system
Step-by-step explanation:
we have
[tex]y\geq x^{2}+x-4[/tex] ----> inequality A
[tex]y < x^{2} +2x+1[/tex] ----> inequality B
we know that
If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities
Verify
For x=0,y=0
Inequality A
[tex]0\geq (0)^{2}+0-4[/tex]
[tex]0\geq -4[/tex] ----> is true
so
The ordered pair satisfy inequality A
Inequality B
[tex]0 < (0)^{2} +2(0)+1[/tex]
[tex]0 < 1[/tex] ----> is true
so
The ordered pair satisfy inequality B
therefore
The ordered pair (0,0) is a solution of the system
(0,0) is a solution to this system y ≥ x² + x - 4 and y < x² + 2x + 1
Discriminant of quadratic equation ( ax² + bx + c = 0 ) could be calculated by using :
From the value of Discriminant , we know how many solutions the equation has by condition :
D < 0 → No Real Roots
D = 0 → One Real Root
D > 0 → Two Real Roots
An axis of symmetry of quadratic equation y = ax² + bx + c is :
[tex]\large {\boxed {x = \frac{-b}{2a} } }[/tex]
Let us now tackle the problem!
Given:
[tex]y \geq x^2 + x - 4[/tex]
[tex]y \leq x^2 + 2x +1[/tex]
[tex]\texttt{ }[/tex]
If we put the coordinate (0 , 0) into the two inequality above then:
[tex]y \geq x^2 + x - 4[/tex]
[tex]0 \geq 0^2 + 0 - 4[/tex]
[tex]0 \geq - 4[/tex] → true
[tex]\texttt{ }[/tex]
[tex]y \leq x^2 + 2x +1[/tex]
[tex]0 \leq 0^2 + 2(0) +1[/tex]
[tex]0 \leq 1[/tex] → true
[tex]\texttt{ }[/tex]
Since the two inequality values are true when the coordinate is (0,0), then it can be concluded that (0,0) is a solution to this system.
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Mathematics
Chapter: Quadratic Equations
Keywords: Quadratic , Equation , Discriminant , Real , Number