How do you solve #\sqrt { 10y ^ { 2} + 23y } = \sqrt { 2y ^ { 2} + 3}#?

1 Answer
Nghi N
Jul 19, 2017

#1/8# and - 3#

Explanation:

Square both sides:
#10y^2 + 23y = 2y^2 + 3#
Bring the equation to standard form:
#f(y) = 8y^2 + 23y - 3 = 0#
Solve it by the new Transforming Method (Google):
Transformed equation:
#f'(y) = y^2 + 23y - 24 = 0#.
Since a +b + c = 0, the 2 real roots of f'(y) are: 1 and #c/a = - 24#
The 2 real roots of f(y) are:
#y1 = 1/a = 1/8# and #y2 = - 24/a = - 24/8 = - 3#

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