How do you solve #4x^2 + 3 = −12x#?

Answer 1

#x = -3/2 +- sqrt6/2#

#y = 4x^2 + 12x + 3 = 0# Use the improved quadratic formula (Google, Yahoo Search) #x = -b/(2a) +- d/(2a).# #D = d^2 = b^2 - 4ac = 144 - 48 = 96# --> #d = +- 4sqrt6# #x = - 12/8 +- 4sqrt6/8# #x = -3/2 +- sqrt6/2#

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Answer 2

Ethan Adkins

To solve the equation 4x^2 + 3 = -12x, first, rewrite it in standard form: 4x^2 + 12x + 3 = 0. Then, use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a, where a = 4, b = 12, and c = 3. Plugging these values into the formula gives: x = (-12 ± √(12^2 - 443)) / (2*4). Simplifying further, x = (-12 ± √(144 - 48)) / 8, x = (-12 ± √96) / 8, x = (-12 ± 4√6) / 8, x = -3/2 ± √6 / 2. So, the solutions are x = (-3 + √6) / 2 and x = (-3 - √6) / 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.