Question

two solutions for all quadratic equations quick check
solve the quadratic equation $10x^2 - 2x + 13 = 3$ which of the following expresses its solutions in the form $a \pm bi$ ? (1 point)
\bigcirc $\frac{1}{10} \pm \frac{\sqrt{101}}{10}i$
\bigcirc $\frac{1}{10} \pm \frac{3\sqrt{11}}{10}i$
\bigcirc $\frac{1}{5} \pm \frac{3\sqrt{11}}{5}i$
\bigcirc $-\frac{1}{10} \pm \frac{3\sqrt{11}}{10}i$

Explanation:

Step1: Simplify the equation

First, we need to rewrite the quadratic equation \(10x^{2}-2x + 13=3\) in standard form \(ax^{2}+bx + c = 0\). Subtract 3 from both sides:
\(10x^{2}-2x+13 - 3=0\)
\(10x^{2}-2x + 10=0\)
We can simplify this equation by dividing all terms by 2:
\(5x^{2}-x + 5=0\) (Wait, actually, dividing \(10x^{2}-2x + 10 = 0\) by 2 gives \(5x^{2}-x+5 = 0\)? No, wait, \(10\div2 = 5\), \(- 2\div2=-1\), \(10\div2 = 5\). But maybe it's better to use the original non - simplified form \(10x^{2}-2x + 10=0\) with \(a = 10\), \(b=-2\), \(c = 10\) for the quadratic formula. The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

Step2: Identify \(a\), \(b\), \(c\)

For the equation \(10x^{2}-2x + 10=0\), we have \(a = 10\), \(b=-2\), \(c = 10\)

Step3: Calculate the discriminant \(\Delta=b^{2}-4ac\)

Substitute the values of \(a\), \(b\), \(c\) into the discriminant formula:
\(\Delta=(-2)^{2}-4\times10\times10\)
\(=4 - 400\)
\(=- 396\)

Step4: Apply the quadratic formula

\(x=\frac{-(-2)\pm\sqrt{-396}}{2\times10}=\frac{2\pm\sqrt{396}i}{20}\) (since \(\sqrt{-396}=\sqrt{396}\times\sqrt{-1}=\sqrt{4\times9\times11}i = 6\sqrt{11}i\))
Simplify \(\frac{2\pm6\sqrt{11}i}{20}\) by dividing numerator and denominator by 2:
\(x=\frac{1\pm3\sqrt{11}i}{10}=\frac{1}{10}\pm\frac{3\sqrt{11}}{10}i\)

Answer:

\(\frac{1}{10}\pm\frac{3\sqrt{11}}{10}i\) (The second option: \(\boldsymbol{\frac{1}{10}\pm\frac{3\sqrt{11}}{10}i}\))