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How do you find the zeros, real and imaginary, of #y=x^2 -x+1# using the quadratic formula?

Answer 1

Eva Adamson

Roots are #= ( 1 + isqrt3)/2, ( 1 - isqrt3)/2#

Standard form of quadratic equation is #ax^2 + bx + c#

Roots are # (-b +- sqrt(b^2 - (4*a*c)))/(2a)#

#y = x^2 -x - 1#

#a = 1, b = -1, c = 1#

Roots are #( 1 +- sqrt(1 - 4))/2#

Roots are #= (1/2)( 1+ isqrt3), (1/2)( 1- isqrt3)#

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Answer 2

Eva Abramson

To find the zeros of ( y = x^2 - x + 1 ) using the quadratic formula, first identify the coefficients ( a = 1 ), ( b = -1 ), and ( c = 1 ). Then apply the quadratic formula:

[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ]

Substitute the values:

[ x = \frac{{-(-1) \pm \sqrt{{(-1)^2 - 4 \cdot 1 \cdot 1}}}}{{2 \cdot 1}} ]

[ x = \frac{{1 \pm \sqrt{{1 - 4}}}}{{2}} ]

[ x = \frac{{1 \pm \sqrt{{-3}}}}{{2}} ]

Since the discriminant ( b^2 - 4ac = -3 ) is negative, the roots are imaginary. Therefore, the zeros of ( y = x^2 - x + 1 ) are complex numbers, given by:

[ x = \frac{{1 \pm \sqrt{{-3}}}}{{2}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.