Question

Determine the number and type of solutions for x2 + 2x + 1= 0.

1 Rational Double

2 Irrational

2 Rational

2 Complex

Answer 1

Well, first, you need to find the solutions :) I'll help you with that :)

Answer 2

What formula do you want to use? >.<

Answer 3

Thank you. :)

Answer 4

Anything is fine.

Answer 5

Lol >.<
it's already in standard form (ax^2 + bx + c = 0)
so a = 1
b = 2
c = 1

fit that into the equation;
\[x = \frac{ -(2) \pm \sqrt{(2)^2 - 4(1)(1)} }{ 2(1) }\]

Answer 6

I see.

Answer 7

\[x = \frac{ -2 \pm \sqrt{4 - 4} }{ 2 }\]

Answer 8

So it is an A?

Answer 9

the answer is irrational

Answer 10

Did you work out the problem? (sorry my laptop shut down on me)

Answer 11

Um.. No.. it's not

Answer 12

yes i did

Answer 13

It's not irrational. Do you know what irrational is?

Answer 14

Both solutions are 1. So x = 1 and x = 1

Answer 15

i put the wrong word i meant rational

Answer 16

So it is x = -+ -2?? @jadeishere

Answer 17

No...the solutions are \[x = 1,~x=1\]

Answer 18

Here's how I would approach this problem.

\[x^2+2x+1 = 0\]For a quadratic such as this, in form \[ax^2 + bx +c = 0\]we can use a quantity called the discriminant (\(\Delta\)) to characterize the solutions.

\[\Delta = b^2-4ac\]

If \(\Delta = 0\) we have a perfect square, with two identical solutions

if \(\Delta > 0\), we have two rational solutions

if \(\Delta < 0\), we have two complex solutions in a complex conjugate pair \(a\pm bi\) where \(i = \sqrt{-1}\)

Doing that for this problem, we see that \[a=1,~b = 2,~c=1\]and \[\Delta = (2)^2-4(1)(1) = 4-4 =0\]so we have a perfect square, or what this problem calls a rational double.

Answer 19

If we solve the problem by factoring or quadratic formula or whatever, we end up seeing that

\[x = \frac{-2\pm\sqrt{(2)^2 - 4(1)(1)}}{2(1)} = \frac{-2\pm\sqrt{0}}{2} = -1\pm 0 = -1\]

Serves me right for not checking the previous answer before quoting it!

Let's check:
\[x^2+2x+1 = 0\]\[(-1)^2 + 2(-1)+1 = 0\]\[1-2+1 = 2-2 = 0\checkmark\]

Answer 20

Omg, thank you. It makes more sense.

Answer 21

If we factor;

\[x^2+2x+1 = 0\]\[(x+1)(x+1) = 0\]\[x+1 = 0\]\[x=-1\]

Answer 22

When you have an even number of repeated roots like we do here, that means instead of crossing the x-axis, the curve just touches and moves away. If we plot \(y = x^2+2x+1\) we get a parabola with a vertex at \((-1,0)\) as you can see in the graph I've attached:

Answer 23

If we didn't have a perfect square, say it was \[x^2+2x-3 = 0\] then we would have two crossings of the x-axis, which would be our solutions. Can you identify them in this graph?

Answer 24

I'm sorry my laptop is being crazy bad.. :/

Answer 25

For that equation, \[\Delta = (2)^2-4(1)(-3) = 4-(-12) = 4+12 = 16\]so we have two distinct, rational roots

\[x = \frac{-2\pm\sqrt{2^2-4(1)(-3)}}{2(1)} =\frac{-2\pm 4}{2} = -1\pm 2 = 1,-3 \]

Answer 26

And that's exactly where we see the parabola crossing the x-axis in the second graph.

Answer 27

Thank you. @whpalmer4 :)

Answer 28

To complete the trifecta, suppose we had \[x^2+2x+2 = 0\]Here the discriminant is \[\Delta = 2^2-4(1)(2) = -4\]so we have two complex roots. The graph shows this by not crossing the x-axis at all!

Answer 29

If we solve, we get
\[x = \frac{-2\pm \sqrt{4-4(1)(2)}}{2(1)} = \frac{-2\pm\sqrt{-4}}{2} = \frac{-2\pm\sqrt{4*(-1)}}{2} = \frac{-2\pm 2i}{2}\]\[\qquad = -1\pm i\]

Answer 30

as promised, they are in the form \(a\pm bi\)

Answer 31

I see.