Question

1. noah and lin are each trying to solve the equation (x^2 - 6x + 10 = 0). they know that the solutions to (x^2 = -1) are (i) and (-i) but they are not sure how to use this information to solve for (x) in their equation.

a. here is noah’s work:

(x^2 - 6x + 10 = 0)

(x^2 - 6x = -10)

(x^2 - 6x + 9 = -10 + 9)

((x - 3)^2 = -1)

show how noah can finish his work using complex numbers.

Explanation:

Step1: Take square root of both sides

Given \((x - 3)^2 = -1\), take the square root of both sides. We know that for any real or complex number \(a\), if \(y^2=a\), then \(y = \pm\sqrt{a}\). So, \(\sqrt{(x - 3)^2}=\pm\sqrt{- 1}\)

Step2: Simplify square roots

We know that \(\sqrt{(x - 3)^2}=\vert x - 3\vert\), but since we are dealing with complex numbers and we can solve for \(x\) directly, we can write \(x - 3=\pm\sqrt{-1}\). And we know that \(\sqrt{-1}=i\) (by the definition of the imaginary unit \(i\)), so \(x - 3=\pm i\)

Step3: Solve for x

Add 3 to both sides of the equation \(x - 3=\pm i\). So, \(x=3\pm i\)

Answer:

To finish Noah's work:

1. Start with \((x - 3)^2=-1\).
2. Take the square root of both sides: \(x - 3=\pm\sqrt{-1}\).
3. Recall that \(\sqrt{-1}=i\) (the imaginary unit), so \(x - 3=\pm i\).
4. Add 3 to both sides: \(x = 3\pm i\).

Thus, the solutions to the equation \(x^{2}-6x + 10 = 0\) are \(x = 3 + i\) and \(x = 3 - i\).