Question

solve
p^2-p+1=0
please show working

Answer 1

(2p-1)(2P-1)=0
p=1/2

Answer 2

how do you get p =1/2

Answer 3

\[p^{2}-p+1=0\]
by using the quadratic formula,
\[p=\frac{ -b \pm \sqrt{b^{2} - 4ac }}{ 2a } \]
where, a=coeff of p^2, b=coeff of p and c=constant term
\[p=\frac{ 1 \pm \sqrt{(-1)^{2} - 4(1)(1) }}{ 2(1) }\]
\[p=\frac{ 1 \pm \sqrt{1-4} }{ 2 }\]
\[p=\frac{ 1 \pm \sqrt{-3} }{ 2 }\]
\[\sqrt{-3}\] is an imaginary number
\[\sqrt{-3} = \sqrt{3}i\]
so, \[p= \frac{ 1 \pm \sqrt{3}i }{ 2 }\]

Answer 4

Thank you

Answer 5

your welcome :)

Answer 6

write a quadratic equation having the given solutions -2,8
please help

Answer 7

here x=-2 and x=8
x=-2
x+2=0
and
x=8
x-8=0
so,
(x+2)(x-8)=0
x^2-6x-16=0 is the equation