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

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

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

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=√(f\left(a\right)),\:\:-√(f\left(a\right))\\\\x=√(-9),\:x=-√(-9)`

`x=√(-9) = √(-1)√(9)=√(9)i=3i\\\\x=-√(-9)=-√(-1)√(9)=-√(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)`