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How do you solve #x^2+4x+2=0# by factoring?

Answer 1

#x^2+4x+2 = (x+2+sqrt(2))*(x+2-sqrt(2))#
therefore #x^2+4x+2=0#
# rarr x=-2-sqrt(2) " or " x=-2+sqrt(2)#

Honestly, it is unlikely that most people would recognize the given factors. This is one situation where you would be better served by using the quadratic formula for roots. #color(white)("XXXX")x=(-b+-sqrt(b^2-4ac))/(2a)#

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Answer 2

Ethan Adkins

To solve the quadratic equation x^2 + 4x + 2 = 0 by factoring, you first rearrange the equation to make it easier to factor. Subtract 2 from both sides to get x^2 + 4x = -2. Then, you factor the quadratic expression on the left-hand side. To do this, you find two numbers that multiply to give the constant term (2) and add to give the coefficient of the linear term (4). In this case, those numbers are 1 and 3. So, you rewrite the middle term as 1x + 3x. Now, you factor by grouping:

x(x + 3) + 1(x + 3) = 0

Now, you can see that both terms have a common factor of (x + 3), so you can factor that out:

(x + 3)(x + 1) = 0

Finally, you set each factor equal to zero and solve for x:

x + 3 = 0 --> x = -3 x + 1 = 0 --> x = -1

So, the solutions to the equation x^2 + 4x + 2 = 0 are x = -3 and x = -1.

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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.