How do you find the zeros, if any, of #y=x^2-3x+1# using the quadratic formula?
The solutions are
Given a quadratic equation
the quadratic formula states that the solutions, if any, are
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To find the zeros of ( y = x^2 - 3x + 1 ) using the quadratic formula, which is ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ), where ( a ), ( b ), and ( c ) are the coefficients of the quadratic equation:
( a = 1 ) ( b = -3 ) ( c = 1 )
Substitute these values into the quadratic formula:
( x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4(1)(1)}}}}{{2(1)}} )
( x = \frac{{3 \pm \sqrt{{9 - 4}}}}{{2}} )
( x = \frac{{3 \pm \sqrt{5}}}{{2}} )
So the zeros are ( x = \frac{{3 + \sqrt{5}}}{{2}} ) and ( x = \frac{{3 - \sqrt{5}}}{{2}} ).
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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.
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.
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.
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.
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