Question

b. lin decides to solve the equation using the quadratic formula. here is her work:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-(6) \pm \sqrt{(-6)^2 - 4(1)(10)}}{2(1)}$
$x = \frac{6 \pm \sqrt{36 - 40}}{2}$
lin knows 36 − 40 is a negative number and isn’t sure what to do next. show how lin can write her solution using $i$.
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Explanation:

Step1: Simplify the radicand

First, calculate the value inside the square root: \(36 - 40=-4\). So the equation becomes \(x = \frac{6\pm\sqrt{- 4}}{2}\).

Step2: Rewrite the square root of negative number

Recall that for a negative number \( - n\) (where \(n>0\)), \(\sqrt{-n}=\sqrt{n}\cdot\sqrt{- 1}\), and by definition, \(\sqrt{-1} = i\). So \(\sqrt{-4}=\sqrt{4}\cdot\sqrt{-1}=2i\).

Step3: Substitute and simplify the fraction

Substitute \(\sqrt{-4} = 2i\) into the equation: \(x=\frac{6\pm2i}{2}\). Then, divide each term in the numerator by 2: \(x = \frac{6}{2}\pm\frac{2i}{2}=3\pm i\).

Answer:

The solutions are \(x = 3 + i\) and \(x=3 - i\) (or written as \(x = 3\pm i\)).