How do you solve #x^2 + 5x + 1 = 0# using the quadratic formula?

Answer 1

#(-5 +- sqrt21)/2#

Use the improved quadratic formula (Google, Yahoo Search). #D = d^2 = b^2 - 4ac = 25 - 4 = 21 # --> #d = +- sqrt21# There are 2 real roots: #x = -b/(2a) +- d/(2a) = -5/2 +- sqrt21/2 = (-5 +- sqrt21)/2#

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Answer 2

Eva Abramson

To solve the quadratic equation ( x^2 + 5x + 1 = 0 ) using the quadratic formula, ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ), where ( a = 1 ), ( b = 5 ), and ( c = 1 ).

Substitute the values of ( a ), ( b ), and ( c ) into the formula: ( x = \frac{{-5 \pm \sqrt{{5^2 - 4 \cdot 1 \cdot 1}}}}{{2 \cdot 1}} )

Calculate the discriminant: ( \text{Discriminant} = b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot 1 = 25 - 4 = 21 )

Apply the square root to the discriminant: ( \sqrt{21} )

Then, the solutions are: ( x = \frac{{-5 + \sqrt{21}}}{{2}} ) and ( x = \frac{{-5 - \sqrt{21}}}{{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.