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How do you solve #6-p^2/8=-4#?

Answer 1

Julia Adams

#p=+-4sqrt5#

#6-p^2/8=-4#

#hArr6+4=p^2/8#

or #p^2/8=10#

or #p^2=80#

or #p^2-80=0#

or #p^2-(sqrt80)^2=0#

or #(p+sqrt80)(p-sqrt80)=0#

i.e. either #p+sqrt80=0# or #p-sqrt80=0#

i.e. either #p=-sqrt80# or #p=sqrt80#

i.e. #p=+-sqrt(4×4×5)=+-4sqrt5#

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Answer 2

Kennedy Adams

To solve the equation ( \frac{6 - p^2}{8} = -4 ), follow these steps:

Multiply both sides of the equation by 8 to eliminate the fraction: ( 8 \times \frac{6 - p^2}{8} = 8 \times (-4) ).
Simplify: ( 6 - p^2 = -32 ).
Add ( p^2 ) to both sides to isolate ( p^2 ): ( 6 - p^2 + p^2 = -32 + p^2 ).
Simplify: ( 6 = -32 + p^2 ).
Add 32 to both sides: ( 6 + 32 = p^2 ).
Simplify: ( 38 = p^2 ).
Take the square root of both sides: ( p = \pm \sqrt{38} ).

Therefore, the solutions to the equation ( \frac{6 - p^2}{8} = -4 ) are ( p = \sqrt{38} ) and ( p = -\sqrt{38} ).

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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.