Solve the quadratic equation 10x2−2x+13=3 . Which of the following expresses its solutions in the form a±bi ?(1 point)

Question

Solve the quadratic equation  10x2−2x+13=3 . Which of the following expresses its solutions in the form  a±bi  ?(1 point)
Responses

110±101√10i
Start Fraction 1 over 10 End Fraction plus or minus Start Fraction Start Root 101 End Root over 10 End Fraction i

15±311√5i
Start Fraction 1 over 5 End Fraction plus or minus Start Fraction 3 Start Root 11 End Root over 5 End Fraction i

−110±311√10i
negative Start Fraction 1 over 10 End Fraction plus or minus Start Fraction 3 Start Root 11 End Root over 10 End Fraction i

110±311√10i

GPT-4o mini · Answer

To solve the quadratic equation $ 10x^2 - 2x + 13 = 3 $, we first rearrange it into standard form: $$ 10x^2 - 2x + 13 - 3 = 0 $$ This simplifies to: $$ 10x^2 - 2x + 10 = 0 $$ Next, we can solve this quadratic equation using the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Here, $ a = 10 $, $ b = -2 $, and $ c = 10 $. Plugging in these values: 1. Calculate $ b^2 - 4ac $: $$ b^2 = (-2)^2 = 4 $$ $$ 4ac = 4 \cdot 10 \cdot 10 = 400 $$ $$ b^2 - 4ac = 4 - 400 = -396 $$ 2. Now plug this into the quadratic formula: $$ x = \frac{-(-2) \pm \sqrt{-396}}{2 \cdot 10} = \frac{2 \pm \sqrt{-396}}{20} $$ 3. Simplify $ \sqrt{-396} $: $$ \sqrt{-396} = \sqrt{396} i = \sqrt{4 \cdot 99} i = 2\sqrt{99} i = 2\sqrt{9 \cdot 11} i = 6\sqrt{11} i $$ 4. Substituting back gives: $$ x = \frac{2 \pm 6\sqrt{11} i}{20} = \frac{2}{20} \pm \frac{6\sqrt{11} i}{20} = \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i $$ Thus, the solutions of the equation are: $$ x = \frac{1}{10} \pm \frac{3\sqrt{11}}{10} i $$ Looking at the provided options, we can see that the correct expression for the solutions in the form $ a \pm bi $ is: $$ \text{Start Fraction 1 over 10 End Fraction plus or minus Start Fraction 3 Start Root 11 End Root over 10 End Fraction i} $$ Therefore, this is the answer.