Question

A: What are the solutions to the quadratic equation x2+9=0? B: What is the factored form of the quadratic expression x2+9? Select one answer for question A, and select one answer for question B. B: (x+3)(x−3) B: (x+3i)(x−3i) B: (x−3i)(x−3i) B: (x+3)(x+3) A: x=3 or x=−3 A: x=3i or x=−3i A: x=3 A: x=−3i

Answer 1

Answer:

A. The solutions are $x=3i,\:x=-3i$.

B. The factored form of the quadratic expression $x^2+9=(x-3i)(x+3i)$

Step-by-step explanation:

A. To find the solutions to the quadratic equation $x^2+9=0$ you must:

$\mathrm{Subtract\:}9\mathrm{\:from\:both\:sides}\\\\x^2+9-9=0-9\\\\\mathrm{Simplify}\\\\x^2=-9\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-9},\:x=-\sqrt{-9}$

$x=\sqrt{-9} = \sqrt{-1}\sqrt{9}=\sqrt{9}i=3i\\\\x=-\sqrt{-9}=-\sqrt{-1}\sqrt{9}=-\sqrt{9}i=-3i$

The solutions are:

$x=3i,\:x=-3i$

B. Two expressions are equivalent to each other if they represent the same value no matter which values we choose for the variables.

To factor $x^2+9$:

First, multiply the constant in the polynomial by $i^2$ where $i^2$ is equal to -1.

$x^2+9i^2$

Since both terms are perfect squares, factor using the difference of squares formula

$a^2-i^2=(a+i)(a-i)$

$x^2+9=x^2+9i^2=\left(-3i+x\right)\left(3i+x\right)$