How do you solve #x^2 +8x +16=0#?

Answer 1

The solution is # x = -4#

#x^ 2 + 8x + 16 = 0#
The equation is of the form #color(blue)(ax^2+bx+c=0# where: #a=1, b= 8, c=16#

The source of the discriminant is:

#Delta=b^2-4*a*c#
# = (8)^2-(4 * 1 * 16)#
# = 64 - 64 = 0 #
The solutions are found using the formula #x=(-b+-sqrtDelta)/(2*a)#
#x = (- 8+-sqrt(0))/(2 * 1 ) = (-8+-0 )/2#
#x = ( - 8 + 0 ) /2 = - 8/2 = - 4#
#x = ( - 8 - 0 ) /2 = - 8/2 = - 4 #
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Answer 2

To solve the equation ( x^2 + 8x + 16 = 0 ), you can use the quadratic formula, which states that for an equation in the form ( ax^2 + bx + c = 0 ), the solutions are given by ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ).

For this specific equation:

  • ( a = 1 )
  • ( b = 8 )
  • ( c = 16 )

Plug these values into the quadratic formula and solve for ( x ):

[ x = \frac{{-8 \pm \sqrt{{8^2 - 4 \cdot 1 \cdot 16}}}}{{2 \cdot 1}} ] [ x = \frac{{-8 \pm \sqrt{{64 - 64}}}}{2} ] [ x = \frac{{-8 \pm \sqrt{{0}}}}{2} ] [ x = \frac{{-8 \pm 0}}{2} ] [ x = \frac{{-8}}{{2}} ]

Thus, the solution is ( x = -4 ) with a multiplicity of 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.

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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