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.

$x^2 - 4 = 0$
$x^2 + 4x - 12 = 0$
$x^2 + 4x + 4 = 0$
$x^2 - 4x + 4 = 0$

a quadratic equation with opposite solutions can be found by multiplying $\boldsymbol{hspace{5em}}$ the equation will have $\boldsymbol{hspace{10em}}$

Explanation:

Step1: Identify opposite-solution quadratics

A quadratic with opposite roots $r$ and $-r$ factors as $(x-r)(x+r)=x^2-r^2$. For $r=2$, this gives $x^2 - 4 = 0$.

Step2: Identify 2-solution quadratics

A quadratic has 2 distinct solutions if its discriminant $b^2-4ac>0$. For $x^2 + 4x - 12 = 0$, discriminant is $4^2-4(1)(-12)=16+48=64>0$. Also $x^2 - 4 = 0$ has discriminant $0^2-4(1)(-4)=16>0$, so it also qualifies.

Step3: Fill the blanks

A quadratic equation with opposite solutions can be found by multiplying $(x - r)(x + r)$. The equation will have no linear ($x$) term.

Answer:

a. $x^2 - 4 = 0$
b. $x^2 + 4x - 12 = 0$ (or $x^2 - 4 = 0$)