How is this equation solved. Have the right to you please help me through this. Ns to the power of 2 minus ns plus 1 equates to 0 p^2-p+1=0 i don"t understand what rather to say to exlain this question. It continues to questioning for an ext detail. Not certain what it wants me come do. Every I require is the equation solved please. Give thanks to you The quadratic formula says that for any quadratic equation wherein ax2 + bx + c = 0, whereby a is the coefficient the the squared term, b is the coefficient that the term elevated to the an initial power, and c is the constant, the variable will equal <-b+/-√(b2 - 4ac)>/(2a).

You are watching: B plus or minus the square root of 4ac

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

Using substitution, we have the right to see the following:

p = <-(-1) +/- √(12 - 4*1*1)>/(2*1)

p = <1 +/- √(1 - 4)>/2

p = <1 +/- √(-3)>/2

If you have not began working through the imaginary number, climate you can conclude the there is no real solution to this equation due to the fact that you cannot take the square root of a negative number. If you have actually started working through the imaginary number, then you can proceed by factoring the out.

p = (1 +/- i√3)/2

p = 1/2 +/- (√3/2)i

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using the quadratic formula will offer you the root (answers) because that the value(s) the x because that this quadratic equation.

as you know, this equation is that a parabola. A parabola has actually 2 roots once it crosses the x axis at 2 points. It has actually one source if it touches the x axis at one point. It has actually NO actual solutions if that 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 the the square term (in her equation that is p2 (who"s coefficient is 1)

"b" is the coefficient of the center term (in your instance it is ns (whos coefficient is additionally 1)

"c" is the coefficient that the third or critical term which is a constant (in your instance it is also 1)

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

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

so there space NO genuine roots, but there ARE facility roots! psychic THAT facility ROOTS constantly COME IN PAIRS called CONJUGATES.

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

In situation you forgot i = sqrt(-1) therefore sqrt(-3) = i(sqrt(3)).

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