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.

How do you solve the system x^2-2x+4+y^2-10=0x^2-2x+4+y^2-10=0 and 2y^2-x+3=02y^2-x+3=0?