How do you solve #y^ { 4} - 10y ^ { 2} = 9#?

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Answer 1 · 2017-05-25T20:49:00

#-3.291#, #3.291#

graph{x^4-10x^2 [-10, 10, -10, 10]}

this is the graph of #f(x)=x^4-10x^2#

the solutions to the equation in question, #y^4-10y^2=9#, are the #x#-coordinates of the points on the graph where #y=9#.

this can be shown by drawing the line #y=9# and finding the points of intersection:
enter image source here

as seen in the image, the points of intersection are #(-3.291,9)# and #(3.291,9).#

the #x#-coordinates are #-3.291# and #3.291# (most likely rounded to #3# decimal places).

therefore the solutions of #y^4-10y^2=9# are #-3.291# and #3.291#.

Answer 2 · 2017-05-28T05:01:36

# y = pmsqrt(5 \pm sqrt(34) ) ~=pm3.291, pm0.912i#

An alternate way to calculate this is to set #Y=y^2#. This gives us:

#y^4-10y^2=9#

#Y^2-10Y=9#

#Y^2-10Y-9=0#

# Y = (-b \pm sqrt(b^2-4ac)) / (2a) #

with #a=1, b=-10, c=-9#

# Y = (10 \pm sqrt((-10)^2-4(1)(-9))) / (2(1)) #

# Y = (10 \pm sqrt(100+36)) / 2 #

# Y = (10 \pm sqrt(136)) / 2 #

# Y = (10 \pm 2sqrt(34)) / 2 #

# Y = 5 \pm sqrt(34) #

and now substitute back in for #y#:

# y^2 = 5 \pm sqrt(34) #

# y = pmsqrt(5 \pm sqrt(34) ) #

#:. => +sqrt(5+sqrt34)~=3.291#
#:. => +sqrt(5-sqrt34)~=0.912i#
#:. => -sqrt(5+sqrt34)~=-3.291#
#:. => -sqrt(5-sqrt34)~=-0.912i#