Question

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

Answer 1

Answer:

Yes, the ordered pair (0,0) is a solution of the system

Step-by-step explanation:

we have

$y\geq x^{2}+x-4$ ----> inequality A

$y < x^{2} +2x+1$ ----> 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

$0\geq (0)^{2}+0-4$

$0\geq -4$ ----> is true

so

The ordered pair satisfy inequality A

Inequality B

$0 < (0)^{2} +2(0)+1$

$0 < 1$  ----> is true

so

The ordered pair satisfy inequality B

therefore

The ordered pair (0,0) is a solution of the system

Answer 2

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

Further explanation

Discriminant of quadratic equation ( ax² + bx + c = 0 ) could be calculated by using :

D = b² - 4 a c

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 :

$\large {\boxed {x = \frac{-b}{2a} } }$

Let us now tackle the problem!

Given:

$y \geq x^2 + x - 4$

$y \leq x^2 + 2x +1$

$\texttt{ }$

If we put the coordinate (0 , 0) into the two inequality above then:

$y \geq x^2 + x - 4$

$0 \geq 0^2 + 0 - 4$

$0 \geq - 4$ → true

$\texttt{ }$

$y \leq x^2 + 2x +1$

$0 \leq 0^2 + 2(0) +1$

$0 \leq 1$ → true

$\texttt{ }$

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.

$\texttt{ }$

Learn more

• Solving Quadratic Equations by Factoring : brainly.com/question/12182022
• Determine the Discriminant : brainly.com/question/4600943
• Formula of Quadratic Equations : brainly.com/question/3776858

Answer details

Grade: High School

Subject: Mathematics

Chapter: Quadratic Equations

Keywords: Quadratic , Equation , Discriminant , Real , Number