How do you find the roots, real and imaginary, of #y=(x+1)(-5x+5)# using the quadratic formula?

Answer 1

Eva Abramson

#x=-1# or #x=1#

We do not Need the quadratic formula, since we get #(x+1)(-5x+5)=0# if #x+1=0# or #-5x+5=0#

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Answer 2

Avery Alexander

To find the roots of ( y = (x+1)(-5x+5) ) using the quadratic formula:

First, expand the expression ( (x+1)(-5x+5) ) to get the quadratic equation in standard form: ( -5x^2 + 5x - 5x - 5 ).
Simplify the quadratic equation: ( -5x^2 - 5 ).
Apply the quadratic formula ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ) to find the roots, where ( a = -5 ), ( b = 0 ), and ( c = -5 ).

[ x = \frac{{-0 \pm \sqrt{{0^2 - 4 \cdot (-5) \cdot (-5)}}}}{{2 \cdot (-5)}} ]

[ x = \frac{{\pm \sqrt{{-100}}}}{-10} ]

[ x = \frac{{\pm 10i}}{{-10}} ]

[ x = \frac{{\pm i}}{{-1}} ]

So, the roots are: ( x = -i ) and ( x = i ).

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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.