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How do you solve #(x + 4)^2 = 11#?

Answer 1

Genesis Allen

#x=-4-sqrt(11)color(white)("XXX")#or#color(white)("XXX")x=-4+sqrt(11)#

Given #color(white)("XXX")(x+4)^2=11#

then #color(white)("XXX")x+4 = +-sqrt(11)# and #color(white)("XXX")x=-4+-sqrt(11)#

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Answer 2

Savannah Anderson

To solve the equation (x + 4)^2 = 11, follow these steps:

Expand the left side of the equation: (x + 4)^2 = (x + 4)(x + 4) = x^2 + 8x + 16
Set the expanded expression equal to 11: x^2 + 8x + 16 = 11
Subtract 11 from both sides: x^2 + 8x + 16 - 11 = 0 x^2 + 8x + 5 = 0
Now, this is a quadratic equation. To solve it, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = 8, and c = 5.
Plug the values into the quadratic formula and simplify: x = (-8 ± √(8^2 - 4(1)(5))) / (2*1) x = (-8 ± √(64 - 20)) / 2 x = (-8 ± √44) / 2
Simplify the square root of 44: √44 = √(4 * 11) = 2√11
Substitute back into the equation: x = (-8 ± 2√11) / 2
Simplify further: x = -4 ± √11

So, the solutions to the equation (x + 4)^2 = 11 are: x = -4 + √11 x = -4 - √11

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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.