solve by using quadratic formula

Question

How is this equation solved.  Can you please help me with this. p to the power of 2 minus p plus 1 equals 0 p^2-p+1=0 I don't know what else to say to exlain this question.  It continues to ask for more detail.  Not sure what it wants me to do.  All I need is the equation solved please.  Thank you

Answer 1

The quadratic formula states that for any quadratic equation where ax² + bx + c = 0, where a is the coefficient of the squared term, b is the coefficient of the term raised to the first power, and c is the constant, the variable will equal [-b+/-√(b² - 4ac)]/(2a).

For the equation that you gave, a = 1, b = -1, and c = 1

Using substitution, we can see the following:

p = [-(-1) +/- √(1² - 4*1*1)]/(2*1)

p = [1 +/- √(1 - 4)]/2

p = [1 +/- √(-3)]/2

If you have not started working with the imaginary number, then you can conclude that there is no real solution to this equation because you cannot take the square root of a negative number.  If you have started working with the imaginary number, then you can continue by factoring it out.

p = (1 +/- i√3)/2

p = 1/2 +/- (√3/2)i

 

Answer 2

using the quadratic formula will give you the roots (answers) for the value(s) of x for this quadratic equation.

as you know, this equation is of a parabola.  A parabola has 2 roots when it crosses the x axis at 2 points.  It has one root if it touches the x axis at one point.  It has NO REAL solutions if it does not cross or touch the x axis.

the quadratic formula     -b plus or minus sqrt(b2-4ac) all divided by 2a.

remember that "a" is the coefficient of the square term (in your equation it is p² (who's coefficient is 1)

"b" is the coefficient of the middle term (in your case it is p (whos coefficient is also 1)

"c" is the coefficient of the third or last term which is a constant (in your case it is also 1)

so we have -1 +- sqrt(1-4) all divided by 2.

thus we have 1 +-i(sqrt(3)) all divided by 2.

so there are NO REAL roots, but there ARE complex roots! REMEMBER THAT COMPLEX ROOTS ALWAYS COME IN PAIRS CALLED CONJUGATES.

THE ROOTS ARE  (1 +i(SQRT(3)))/2 AND (1-i(SQRT(3)))/2.

In case you forgot i = sqrt(-1) so sqrt(-3) = i(sqrt(3)).

I know this looks ugly because I do not have use of all the symbols on this answer.

If you need more help, let me know.

 

Stu