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
$x^2 - 4 = 0$
$x^2 + 4x - 12 = 0$
$x^2 + 4x + 4 = 0$
$x^2 - 4x + 4 = 0$
$(x - r)$ and $(x + r)$
$(x - r)$ and $(x - r)$
$(x + r)$ and $(x + r)$
a quadratic equation with opposite solutions can be found by multiplying ▼ the equation will have ▼
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Explanation:

Step1: Analyze opposite solutions

If solutions are opposites, let them be $r$ and $-r$. The factored quadratic is $(x-r)(x+r)$.

Step2: Expand the factored form

$$(x-r)(x+r) = x^2 - r^2$$
For $r=2$, this gives $x^2 - 4 = 0$, with solutions $2$ and $-2$ (opposites).

Step3: Identify two-solution quadratic

A quadratic with two distinct real solutions has a positive discriminant. For $x^2 + 4x - 12 = 0$, discriminant $\Delta = 4^2 - 4(1)(-12) = 16 + 48 = 64 > 0$, so it has two distinct solutions.

Answer:

a. Quadratic equation: $x^2 - 4 = 0$; Factored form source: $(x-r)$ and $(x+r)$; Solution property: two distinct real opposite solutions
b. Example quadratic equation: $x^2 + 4x - 12 = 0$