Question

If 2+i and 2+i are the roots of the equation x^2-4x+c=0, what is value of c?

Answer 1

You can find the answer a couple of ways. One would be to rely on the fact that if \(a\) is a root of the polynomial, then evaluating the polynomial at \(x = a\) should give you 0.

Another is to recall that if you factor a polynomial and set it equal to 0, the factors = 0 at the roots.

Answer 2

You could plug in one of the values for \(x\), expand the equation and solve for \(c\).

Answer 3

Are you asking for help with another question, or what?

Answer 4

both @whpalmer4

Answer 5

Do you have an answer for the first question yet?

Answer 6

i got c

Answer 7

Please, you couldn't know this, but people who answer my questions about the answer with the letter of the answer choice drive me nuts. I don't care what the stinking answer choice is, I want to know the answer to the equation...

Answer 8

-4

Answer 9

alas, answer choice c is incorrect. Show me your work and we'll figure out where you went wrong.

Answer 10

by the way, you copied the problem incorrectly, but I misread it correctly :-)

Answer 11

sometimes two wrongs do make a right :-)

Answer 12

oh i see where i went wrong i checked it over its suppose to be positive so it would be 4

Answer 13

Alas, incorrect again...

Answer 14

Here's how you test your answer, to verify that it is correct. Let's say we think c = 4:

\[x^2-4x+4=0\]Choose one of the roots, how about \(x = 2+i\)
\[(2+i)^2-4(2+i) +4=0\]\[2^2+2i+2i + i^2 - 4*2 -4*i + 4 = 0\]\[4 + 4i + i^2 - 8 - 4i + 4 - 0\]\[i^2=0\]Oops.

Answer 15

\[i = \sqrt{-1}\rightarrow i^2 = -1\]So \[i^2\ne0\]

Answer 16

wow, i guess i just don't understand this

Answer 17

You don't understand my check, or how to solve the problem? I guess both is also an allowable choice :-)

Answer 18

both, but its okay i have more problems to do ill get back to it, thank you for your help

Answer 19

Roots of \[x^2-4x+4=0\] are \(x=2,\, x=-2\), btw .

If you aren't getting this, you might well get the others wrong, too. Stick around!

Answer 20

Let's just plug one of the roots into the equation and see what \(c\) must equal.

\[x^2-4x+c=0\]\[(2+i)^2 -4(2+i) + c = 0\]\[(2*2 + 2*i + 2*i + i*i) - 4*2 - 4*i + c = 0\]\[4+4i + i^2 - 8 - 4i + c = 0\]Agree so far?

Answer 21

yes

Answer 22

Okay, simplify that last equation for me, what do you get?

Answer 23

4+4i+i2−8−4i+c=0

i2=2

Answer 24

no, what happened to c?

Answer 25

\[4+4i+i^2-8-4i+c=0\]\[4\cancel{+4i}+i^2-8\cancel{-4i}+c=0\]
\[4+i^2-8+c=0\]\[-4+i^2+c=0\]\[c=4-i^2\]\[c=\]

Answer 26

4-( -1^2) translation right

Answer 27

no, \[i^2=-1\]
\[c = 4-(-1)\]\[c=5\]

Let's test our answer with the other root:

\[x^2-4x+5=0\]\[(2-i)^2-4(2-i)+5=0\]\[(2*2-2*i-2*i-i(-i)) -4*2-4*(-i) + 5 = 0 \]\[4-4i+i^2-8+4i+5=0\] \[1+i^2=0\]\[1+(-1)=0\checkmark\]

Answer 28

So for the second problem, do you know how to find the solutions for those equations?

Answer 29

no, my teacher just gave us this stuff , I'm still in algebra, but he's giving us trig work and he's not teaching us anything, so I'm lost

Answer 30

This isn't trig, this is algebra. Have you learned factoring, or the quadratic formula?

Answer 31

none at all, all we did so far was basics and we have regents to take in june

Answer 32

Well, do you want to learn them now? :-)

Answer 33

yes

Answer 34

Okay, let's do the quadratic formula first, which will at least allow you to make progress on this problem while I write up a description of factoring.

If you have a quadratic polynomial in one variable of the form \[ax^2+bx+c = 0,\,\,a\ne0\]Solutions for the quadratic will be given by \[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]The quantity under the radical sign is called the discriminant, sometimes written as \(\Delta = b^2-4ac\) and gives us a convenient way to predict the form of the roots.

If \(\Delta = 0\) we have a "perfect" square, and will have just one value for \(x\) (*)
If \(\Delta >0\) we will have two different real values for \(x\)
If \(\Delta < 0 \) we will have two complex values for \(x\) and they will make what is called a complex conjugate pair: \(a\pm bi\)

(*) While the root only has one value, it has a multiplicity of 2, meaning it is a repeated root that appears twice in the equation.

Answer 35

You can use that formula to find the roots of all 4 equations in the problem. I think it will be obvious when you solve one of them that you've found the right answer, and wouldn't bother trying to multiply the roots together as you find them.

Answer 36

what does ± mean?

Answer 37

just what it looks like: plus or minus. It's a shorthand for this:

\[x = \frac{-b+\sqrt{b^2-4ac}}{2a},\,\,x = \frac{-b-\sqrt{b^2-4ac}}{2a}\]

Answer 38

Here's an example:
\[x^2+2x+1 = 0\]\[a=1,b=2,c=1\]\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-2\pm\sqrt{2^2-4(1)(1)}}{2*1} = \frac{-2\pm\sqrt{4-4}}{2} = -1\]Plugging in to the original equation to test it:
\[(-1)^2+2(-1)+1=0\]\[1-2+1=0\checkmark\]This was a case where \(\Delta=0\) so we have one answer only.

Answer 39

Let's try another:
\[x^2+3x+2=0\]\[a=1,b=3,c=2\]\[x=\frac{-3\pm\sqrt{(3)^2-4(1)(2)}}{2(1)} = \frac{-3\pm\sqrt{9-8}}{2} = \frac{-3\pm1}{2}\]\[x=-2, \,\,x = -1\]are the solutions:
\[(-2)^2+3(-2)+2=0\]\[4-6+2=0\]\[0=0\checkmark\]\[(-1)^2+3(-1)+2=0\]\[1-3+2=0\]\[0=0\checkmark\]
Here we had \(\Delta > 0\) so we got two real roots.

Answer 40

Finally, an example with complex roots:\[x^2-2x+2=0\]\[a=1,b=-2,c=2\]\[x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(2)}}{2(1)} = \frac{2\pm\sqrt{ -4}} {2} = \frac{2\pm 2i}{2} =1\pm i\]

Checking the solution:

\[({1+i})^2 -2(1+i)-2=0\]\[(1*1+1*i + 1*i + i*i) - 2*1-2*i +2 = 0\]\[1+2i+i^2-2-2i+2=0\]\[1+i^2=0\]\[1+(-1)=0\]\[0=0\checkmark\]