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How do you solve #x^2 + 8x + 13 = 0#?

Answer 1

Avery Alexander

#x=-4+sqrt 3, -4-sqrt 3#

#x^2+8x+13=0# is a quadratic equation in standard form #ax^2+bx+13#, where #a=1, b=8, c=13#.

The equation can be resolved using the quadratic formula.

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

Substitute the known values into the formula and solve for #x#.

#x=(-8+-sqrt(8^2-(4*1*13)))/(2*1)#

Simplify.

#x=(-8+-sqrt(64-52))/2#

Simplify.

#x=(-8+-sqrt 12)/2#

Factor #sqrt 12#.

#sqrt(2xx2xx3)=#

#sqrt(2^2xx3)=#

#2sqrt 3#

#x=(-8+-2sqrt3)/2#

Simplify.

#x=(cancel(-8)^-4+-cancel(2)^1sqrt 3)/cancel(2)^1#

Simplify.

#x=-4+-sqrt 3#

Solutions for #x#.

#x=-4+sqrt 3#

#x=-4-sqrt 3#

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Answer 2

Nathan Axel

To solve the equation x^2 + 8x + 13 = 0, you can use the quadratic formula. First, identify the coefficients a, b, and c in the quadratic equation ax^2 + bx + c = 0. Then, substitute these values into the quadratic formula: x = [-b ± √(b^2 - 4ac)] / (2a). Calculate the discriminant (the part inside the square root): b^2 - 4ac. Finally, substitute the values of a, b, and c into the quadratic formula and simplify to find the solutions for x.

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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.