How do you solve #8x^2+4=-33x#?

2 Answers
Don't Memorise
Apr 5, 2016

The solutions are:
#x = color(blue)(-1/8#

#x= color(blue)( - 4#

Explanation:

#8x^2 + 4 = -33x#

#8x^2 + 4 + 33x = 0#

#8x^2 + 33x + 4 = 0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:

#a=8, b=33, c=4#

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (33)^2-(4* 8 * 4)#

# = 1089 -128= 961#

The solutions are found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

#=(-33+-sqrt961)/(2*8)#

#=(-33+-31)/(16)#

#x =(-33+31)/(16) = -2/16 = color(blue)(-1/8#

#x =(-33-31)/(16) = -64/16 = color(blue)( - 4#

Nghi N.
Apr 6, 2016

1/8 and 4

Explanation:

#y = 8x^2 + 33x + 4 = 0#
Use the new Transforming Method (Google and Yahoo Search).
Transformed equation: #y' = x^2 + 33x + 32#.
Since a - b + c = 0, use shortcut. Two real roots of y' are: -1 and (-c/a = -32).
Back to y, divide these real roots by (a = 8) to get the 2 real roots of y --> : #x1 = -1/8# and #x2 = -32/8 = -4.#

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