OpenStudy (anonymous):

Factor completely: x^2 − 3x + 5

3 years ago
OpenStudy (anonymous):

@iambatman

3 years ago
OpenStudy (xguardians):

I think this is unfactorable?

3 years ago
OpenStudy (anonymous):

which is prime?

3 years ago
OpenStudy (anonymous):

\[x^2-3x+5 = 0 \] Complete the square \[x^2-3x = -5\] \[x^2-3x+\frac{ 9 }{ 4 }=-\frac{ 11 }{ 4 }\] \[(x-\frac{ 3 }{ 2 })^2=-\frac{ 11 }{ 4 }\] Eliminate the exponents now, and you'll get complex answers.

3 years ago
OpenStudy (anonymous):

cuz i was thinking it was prime idk

3 years ago
OpenStudy (anonymous):

is that considered prime or?

3 years ago
OpenStudy (rishavraj):

... u wont be able to factorise tht

3 years ago
OpenStudy (anonymous):

Like I said you'll get complex roots.

3 years ago
OpenStudy (rishavraj):

\[x = (3\pm \sqrt{11}i)/2\]

3 years ago