How do you solve the system $x^{2} + y^{2} - 3 x = 8$ $x^2+y^2-3x=8$ and $2 x^{2} - y^{2} = 10$ $2x^2-y^2=10$ ?

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Answer 1 · 2016-08-24T05:53:46

(1) :In $RR^2$ , the Solns. are $(x,y)=(3,+-2sqrt2)$ .

(2) : In $CC^2$ , the Solns. are $(x,y)=(3,+-2sqrt2),(-2, +-isqrt2)$ .

Adding the given eqns., we get, $3x^2-3x=18, or, x^2-x-6=0$ .

$:. ul(x^2-3x)+ul(2x-6)=0$ .

$:. x(x-3)+2(x-3)=0$ .

$:. (x-3)(x+2)=0$ .

$:. x=3, or, x=-2$ .

From the second eqn., since, $y^2=2x^2-10$ ,

$x=3 rArr y^2=8 rArr y=+-2sqrt2$ , and,

$x=-2 rArr y^2=-2, "which is not possible in" RR$ .

These roots satisfy the given system of eqns.

Hence, the Solns. in $RR^2$ are $(x,y)=(3,+-2sqrt2)$ .

Note :-

In $CC, y^2=-2 rArr y=+-isqrt2$ .

Therefore, in $CC^2$ , the Solns. are,

$(x,y)=(3,+-2sqrt2), (-2, +-isqrt2)$ .

Enjoy Maths.!

How do you solve the system x^2+y^2-3x=8x2+y2−3x=8x^2+y^2-3x=8 and 2x^2-y^2=102x2−y2=102x^2-y^2=10?