Question

write a quadratic equation for each condition below. explain your reasoning. a. the equation has solutions that are opposites. b. the equation has two solutions. a. which of the following quadratic equations has solutions that are opposites? a. $x^2 + 4x + 4 = 0$ b. $x^2 + 4x - 12 = 0$ c. $x^2 - 4x + 4 = 0$ d. $x^2 - 4 = 0$

Explanation:

Part 1: Writing quadratic equations

Step1: Opposite solutions equation

For opposite solutions $r$ and $-r$, sum of roots is $r + (-r) = 0$. Using Vieta's formula, the coefficient of $x$ is $-(sum) = 0$. A simple equation is:
$x^2 - 4 = 0$
Reasoning: The roots are $2$ and $-2$ (opposites), and the linear term ($x$-term) has a coefficient of 0, which ensures the sum of roots is 0.

Step2: Two distinct solutions equation

A quadratic equation has two distinct solutions when its discriminant $b^2-4ac>0$. A simple example:
$x^2 - 5x + 6 = 0$
Reasoning: The discriminant is $(-5)^2 - 4(1)(6) = 25-24=1>0$, so it has two distinct roots ($2$ and $3$).

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Part 2: Multiple choice question

For a quadratic equation $ax^2+bx+c=0$, solutions are opposites if the sum of roots $\frac{-b}{a}=0$, meaning $b=0$. We check each option:

• A. $x^2+4x+4=0$: $b=4

eq0$, roots are $-2$ (repeated, not opposites)

• B. $x^2+4x-12=0$: $b=4

eq0$, roots are $2$ and $-6$ (not opposites)

• C. $x^2-4x+4=0$: $b=-4

eq0$, roots are $2$ (repeated, not opposites)

• D. $x^2-4=0$: $b=0$, roots are $2$ and $-2$ (opposites)

Answer:

D. $x^2 - 4=0$

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Final Combined Answers:

1. a. Quadratic equation with opposite solutions: $\boldsymbol{x^2 - 4 = 0}$ (reasoning: linear term coefficient is 0, sum of roots is 0, giving opposite solutions)
2. b. Quadratic equation with two solutions: $\boldsymbol{x^2 - 5x + 6 = 0}$ (reasoning: discriminant is positive, so two distinct real roots)
3. Multiple choice: D. $x^2 - 4=0$