Question

find the solution(s) of the system of equations.
\\(y = x^2 + 4x\\)
\\(y + x^2 = -4x\\)

a) \\((0, 0)\\) and \\((4, 0)\\)
b) \\((-4, 0)\\) and \\((4, 0)\\)
c) \\((0, 0)\\)
d) \\((-4, 0)\\) and \\((0, 0)\\)

question 5 (5 points)
find all solution(s) of the quadratic equation \\(x^2 - 100 = 0\\).

a) \\(x = 100, -100\\)
b) \\(x = 10\\)
c) \\(x = 10, -10\\)
d) \\(x = -10\\)

question 6 (5 points)
find the solution(s) of the system of equations.
\\(y = -x^2 + 2\\)
\\(y = x^2 + 4x + 4\\)

a) \\((-1, 1)\\) and \\((1, 1)\\)
b) \\((1, 1)\\)
c) \\((-1, 1)\\)
d) \\((-1, 1)\\) and \\((0, 2)\\)

Explanation:

Solve the first system of equations

Substitute \(y = x^2 + 4x\) into the second equation \(y + x^2 = -4x\):

$$ (x^2 + 4x) + x^2 = -4x \implies 2x^2 + 8x = 0 \implies 2x(x + 4) = 0 $$

The solutions for \(x\) are \(x = 0\) and \(x = -4\).

• For \(x = 0\): \(y = 0^2 + 4(0) = 0\). Point: \((0, 0)\).
• For \(x = -4\): \(y = (-4)^2 + 4(-4) = 0\). Point: \((-4, 0)\).

The solutions are \((-4, 0)\) and \((0, 0)\).

Solve the quadratic equation

Solve \(x^2 - 100 = 0\):

$$ x^2 = 100 \implies x = \pm \sqrt{100} \implies x = 10, -10 $$

Solve the second system of equations

Equate the two expressions for \(y\):

$$ -x^2 + 2 = x^2 + 4x + 4 \implies 2x^2 + 4x + 2 = 0 \implies 2(x^2 + 2x + 1) = 0 \implies 2(x+1)^2 = 0 $$

This yields \(x = -1\).

• For \(x = -1\): \(y = -(-1)^2 + 2 = -1 + 2 = 1\). Point: \((-1, 1)\).

The solution is \((-1, 1)\).

Answer:

Question 1

• A) (0, 0) and (4, 0)
• B) (-4, 0) and (4, 0)
• C) (0, 0)
• D) (-4, 0) and (0, 0) (Correct answer)

Question 2

• A) x = 100, -100
• B) x = 10
• C) x = 10, -10 (Correct answer)
• D) x = -10

Question 3

• A) (-1, 1) and (1, 1)
• B) (1, 1)
• C) (-1, 1) (Correct answer)
• D) (-1, 1) and (0, 2)