Home > Algebra > Use Square Roots to Solve Quadratic Equations

How do you solve #8x^2-5=-4x#?

Answer 1

Eva Abramson

#=> x approx 0.57915 #

# x approx -1.07915#

#8x^2-5=-4x#

#=> 8x^2-5+4x=0 #

#=> 8x^2+4x -5=0 # --------(1)

using quadratic equation formula:

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

Substitute the corresponding values of a,b, and c from (1)

#=> x= (-4+- sqrt(4^2-4xx8xx(-5)))/(2xx8)#

#=> x= (-4+- sqrt(16+ 160))/(16)#

#=> x= (-4+- sqrt(176))/(16)#

#=> xapprox -4/16 +- 13.266499/16# ---truncating value of #sqrt(176)#

# =>xapprox -1/4 +- 13.2665/16#

#=> x approx - 0.25 +- 0.82915#

#=> x approx - 0.25 + 0.82915 or x approx - 0.25 - 0.82915#

#=> x approx 0.57915 #

# x approx -1.07915#

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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.