Question

Find x^2-2x-4=0 using the quadratic formula.

I keep getting the wrong answer. I know what the right answer is.

Answer 1

\[x = 1 \pm \sqrt{5} \]

Answer 2

\[x \approx 3.2360679775, ~-1.2360679775\]

Answer 3

How did you get that answer? I keep getting \[\pm2\sqrt{5}\]

Answer 4

Can you show your work? @Rachella

Answer 5

Step One - Find coefficients a, b and c.

\[a = 1, ~b= -2,~c=-4\]
Step Two - Plug in the values for a, b and c into the quadratic formula.

\[x1,2 = \frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a}\]
\[x1,2 = \frac{ -2\pm \sqrt{(-2^{2}-4\times1\times(-4)} }{ 2\times1}\]
Step Three - Simplify expression under the square root.

\[x1,2 = \frac{ 2\pm \sqrt{20} }{ 2 }\]

Step Four - Solve for x.

\[x1 = \frac{ 2+ \sqrt{20} }{ 2 }~= 1 + \sqrt{5}\]
\[x2 = \frac{ 2- \sqrt{20} }{ 2 } = 1 - \sqrt{5}\]

Answer 6

\[Equation~To~Solve: x ^{2} - 2x - 4 = 0 \]

\[The~Solutions~Are: x1 = 1 + \sqrt{5}~ and ~x2= 1 - \sqrt{5}\]

Answer 7

How does the \[\sqrt{20}\] equal \[\sqrt{5}\]

Answer 8

The square root of 20 would be 4.472 which rounded would be 5.

Answer 9

When I simplified \[\sqrt{20}\] , I got \[2\sqrt{5}\]

Answer 10

I agree with @Rachella

Answer 11

I looked in the back of the book and Rachella's answer is correct but I am not sure how she got there....
@Rachella @iKayla

Answer 12

I explained step by step how I got that answer.

Answer 13

hi

Answer 14

How did you go from \[\frac{ 2+\sqrt{20} }{ 2 }\] to \[1 + \sqrt{5}\]

Answer 15

\[\sqrt{20} = \sqrt{4 \times 5} = \color{green}{2 \sqrt{5}}\]

Answer 16

Take out \(2\) common from Numerator and cancel it with \(2\) in denominator.

Answer 17

\[\frac{2 + 2 \sqrt{5}}{2} \implies \frac{\cancel{2}(1+\sqrt{5})}{\cancel{2}} = 1 + \sqrt{5}\]

Answer 18

Thanks so much! It makes more sense now! @waterineyes :)

Answer 19

you are welcome dear.. :)

Answer 20

You basically factored out the two right? @waterineyes

Answer 21

Taking something common means yes factoring out..

I have factored out \(2\) there.

Answer 22

Ok! Thanks again! :)