How do you solve #x(x+6)=-2# using the quadratic formula?

Answer 1

When given an equation of the form, #ax^2+bx+c = 0#, one can find the value(s) of x by substituting the values for a, b, and c into the quadratic formula:

#x = (-b+-sqrt(b^2-4(a)(c)))/(2a)#

The given equation #x(x+6)=-2# is not in the form specified in the answer, therefore, we must put it in that form.

Use the distributive property on the left side:

#x^2 + 6x = -2#

Add 2 to both sides:

#x^2 + 6x +2= 0#

By observation, #a = 1, b = 6, and c = 2#

Substitute these values into the quadratic formula:

#x = (-6+-sqrt(6^2-4(1)(2)))/(2(1))#

#x = (-6+-sqrt(36-8))/2#

#x = (-6+-sqrt(28))/2#

#x = (-6+-2sqrt7)/2#

#x = -3+-sqrt7#

#x = -3-sqrt7# and #x = -3+sqrt7#

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Answer 2

Jameson Allen

To solve the equation x(x + 6) = -2 using the quadratic formula, first, rewrite the equation in the form ax^2 + bx + c = 0, where a = 1, b = 6, and c = 2.

Then, substitute these values into the quadratic formula: x = [-b ± √(b^2 - 4ac)] / (2a).

After substituting the values, solve for x by performing the arithmetic operations within the formula.

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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.