OpenStudy (anonymous):

what are the solution of x^2-4x+5=0? anyone could please help

3 years ago
OpenStudy (solomonzelman):

imaginary numbers

3 years ago
OpenStudy (solomonzelman):

subtract 1 from both sides, and try to complete the square.

3 years ago
OpenStudy (anonymous):

i just started this idk anything

3 years ago
OpenStudy (solomonzelman):

What course is this, ALGEBRA 1 ?

3 years ago
OpenStudy (solomonzelman):

If so, say no solutions.

3 years ago
OpenStudy (yanasidlinskiy):

\(\Huge\bf \color{yellow}{Welcome~to~OpenStudy!!}\hspace{-310pt}\color{cyan}{Welcome~to~OpenStudy!!}\hspace{-307.1pt}\color{midnightblue}{Welcome~to~\color{purple}{Open}}\color{blue}{Study!!!!}\) @SolomonZelman Can teach you!!:)

3 years ago
OpenStudy (anonymous):

algebra 2

3 years ago
OpenStudy (mokeira):

\[x ^{2}-4x+5\] find two numbers which when you multiply give you 5 and when you add give you -4

3 years ago
OpenStudy (solomonzelman):

\(\normalsize\color{blue}{ x^2-4x+5=0}\) \(\normalsize\color{blue}{ x^2-4x+5 \color{red}{-1}=0\color{red}{-1}}\) \(\normalsize\color{blue}{ x^2-4x+4=-1}\) \(\normalsize\color{blue}{ (x-2)^2=-1}\) \(\normalsize\color{blue}{ x-2=\sqrt{-1}}\)

3 years ago
OpenStudy (solomonzelman):

For you, it will be no solutions, or err imput. (on an imaginary level though.... \(\normalsize\color{blue}{ x-2=± \sqrt{-1}}\) \(\normalsize\color{blue}{ x-2=±i}\) \(\normalsize\color{blue}{ x=2±i}\)

3 years ago
OpenStudy (mokeira):

nyc method @SolomonZelman

3 years ago
OpenStudy (solomonzelman):

Yes, I am kind of completing the square, but I am skipping the 1st step of setting it to ax²+bx=-c, I just start with ax²+bx+c=d, and add whatever number that would make it a perf. square.

3 years ago
OpenStudy (anonymous):

so how bout for this one

3 years ago
OpenStudy (anonymous):

x^2-2x+37=0?

3 years ago
OpenStudy (solomonzelman):

no solution subtract 36 from both sides

3 years ago
OpenStudy (solomonzelman):

but, you will get (imaginary solutions) x= -1±6i

3 years ago
OpenStudy (mokeira):

why 36...why not 1? do you subtract a number that is also a perfect square?

3 years ago
OpenStudy (solomonzelman):

Yes, that is what I do, I make the left hand side into a perfect square.

3 years ago
OpenStudy (solomonzelman):

it is completing the square, without a step. My teacher forgave me that :)

3 years ago
OpenStudy (anonymous):

x=1+6i x=1-6i ?

3 years ago
OpenStudy (anonymous):

or x=6i x=-6i

3 years ago
OpenStudy (solomonzelman):

yes, -1+6i, or -1-6i If you haven't learned about imaginary numbers yet, don't say this though. Then, just say no solution.

3 years ago
OpenStudy (anonymous):

okay thanks

3 years ago
OpenStudy (anonymous):

x^2-3x+3=0?

3 years ago
OpenStudy (solomonzelman):

Anytime

3 years ago
OpenStudy (solomonzelman):

x²-3x+3=0 use the quadratic formula, or complete the square. (you can't factor it, because the discriminant isn't a perfect square)

3 years ago
OpenStudy (solomonzelman):

Can you finish it, or need help ?

3 years ago
OpenStudy (anonymous):

i tried it but i can't

3 years ago
OpenStudy (solomonzelman):

\(\LARGE\color{blue}{ x= \frac{-(-3)± \sqrt{(-3)^2-4(1)(3)}}{2(1)} }\) \(\LARGE\color{blue}{ x= \frac{3± \sqrt{9-12}}{2} }\) \(\LARGE\color{blue}{ x= \frac{3± \sqrt{-3}}{2} }\)

3 years ago
OpenStudy (solomonzelman):

So, no real number solutions, but x= (3-i√3) / 2 or (3+i√3) / 2

3 years ago
OpenStudy (solomonzelman):

sorry for a late reply.

3 years ago
OpenStudy (anonymous):

got you

3 years ago
OpenStudy (anonymous):

the discriminant of a quadratic equation is negative. one solution is 3+4i. what is the other solution ?

3 years ago
OpenStudy (solomonzelman):

3+4i is not one of the solutions, I posted the solutions and the way to obtain them. You got right though, that negative discriminant = imaginary solution(s) discriminant 0 = 1 real solutions (or saying that the equation is really a line) positive discriminant = 2 real solutions.

3 years ago