How do you solve # x^2+4x-5=0 #?

Answer 1

#x= -5 or +1#

First, factor this equation. Then, determine what two values add to equal to the middle term #(+4)# and multiply to the last term #(-5)#,
#+5 and -1#

Plug it in the equation:

#(x+5)(x-1)#

and set equal to zero

#(x+5)(x-1)=0#
#x+5=0 => x=-5 #
#x-1=0 => x=+1#
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Answer 2

To solve the quadratic equation ( x^2 + 4x - 5 = 0 ), you can use the quadratic formula or factorization method.

Using the quadratic formula: [ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ]

Where:

  • ( a = 1 )
  • ( b = 4 )
  • ( c = -5 )

Substitute the values into the formula: [ x = \frac{{-4 \pm \sqrt{{4^2 - 4(1)(-5)}}}}{{2(1)}} ] [ x = \frac{{-4 \pm \sqrt{{16 + 20}}}}{{2}} ] [ x = \frac{{-4 \pm \sqrt{{36}}}}{{2}} ] [ x = \frac{{-4 \pm 6}}{{2}} ]

This gives two solutions: [ x_1 = \frac{{-4 + 6}}{{2}} = \frac{2}{2} = 1 ] [ x_2 = \frac{{-4 - 6}}{{2}} = \frac{{-10}}{2} = -5 ]

So, the solutions to the equation ( x^2 + 4x - 5 = 0 ) are ( x = 1 ) and ( x = -5 ).

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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.

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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