How do you solve the system #x^2-2x+4+y^2-10=0# and #2y^2-x+3=0#?

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Answer 1 · 2016-08-25T10:53:12

#x=1/4(3+sqrt129), and, y=+-sqrt{1/8(sqrt129-9)# are the solns.

From the second eqn., we have, #y^2=(x-3)/2#

We subst. this in the first eqn., and , get,

#x^2-2x+4+(x-3)/2-10=0#

#:. 2x^2-4x+8+x-3-20=0#, i.e.,

#2x^2-3x-15=0#

To solve this Quadr. Eqn., we use the Formula : The Roots, #alpha,#

#beta#, of the General Quadr. Eqn. : #ax^2+bx+c=0# are given by,

#alpha, beta =(-b+-sqrt(b^2-4ac))/(2a)#

Applying this Formula, with #a=2, b=-3, c=-15#, we have,

#alpha, beta=(3+-sqrt(9+120))/4=(3+-sqrt129)/4#

Then, since, y^2=(x-3)/2, we get,

#y^2=1/2((3+-sqrt129)/4-3)=1/8(3+-sqrt129-12)=1/8(+-sqrt129-9)#

But, in #RR, y^2=1/8(-sqrt129-9)<0# is not possible.

Hence, #x=1/4(3+sqrt129), and, y=+-sqrt{1/8(sqrt129-9)# are the solns.