Mathematics

# how do you solve this equation? (x-4)+y2=4 x2-8x+y2=-12

#### yassautumn

4 years ago

$(x-4)+y^2=4\\ x^2-8x+y^2=-12\\\\ y^2=4-(x-4)=4-x+4=-x+8\\ x^2-8x+y^2=-12\\\\ x^2-8x-x+8=-12\\ x^2-9x+20=0\\ x^2-4x-5x+20=0\\ x(x-4)-5(x-4)=0\\ (x-5)(x-4)=0\\ x=5 \vee x=4\\\\ y^2=-5+8 \vee y^2=-4+8\\ y^2=3 \vee y^2=4\\ y=-\sqrt3 \vee y=\sqrt3 \vee y=-2 \vee y=2\\\\ x=5,y=-\sqrt3\\ x=5,y=\sqrt3\\ x=4,y=-2\\ x=4,y=2$

#### dalmayshuns101

4 years ago

$(x-4) + y^2 = 4$
$x^2 - 8x + y^2 = 12$$x - 4 + y^2 = 4$
$x + y^2 = 8$$x^2 - 8x + y^2 = -12$
﻿-          x + $y^2$ = 8
$x^2 - 9x = -20$$x^2 - 9x + 20 = 0$
$(x - 5)(x - 4)$
x - 5 = 0           x - 4 = 0
x = 5                x = 4
$x + y^2 = 8$$5 + y^2 = 8$
$y^2 = 3$
$\sqrt{y^2} = \sqrt{3}$
y ≈ +/- 1.73$4 + y^2 = 8$
$y^2 = 4$
$\sqrt{y^2} = \sqrt{4}$
y = +/- 2