Solutions To Mathematics Textbooks/Basic Mathematics/Chapter 8

Chapter 8

Complete the square of the equations in this part to render them in a form of the equation of a circle.

$x^2 + 2x + y^2 = 5 \,$

$x^2 + 2x = 5 - y^2 \,$

$x^2 + 2x + 1 = 6 - y^2 \,$

We add one to both sides so we can write the right hand side as a single power.

$(x+1)^2 = 6 - y^2 \,$

$(x+1)^2 + y^2 = 6 \,$

$x^2 + 4x + y^2 - 4y = 20 \,$

$x^2 + 4x = -(y^2-4y) + 20 \,$

$x^2 + 4x + 4 = -(y^2-4y)+20 \,$

$(x+2)^2 = -(y^2-4y)+24 \,$

$y^2-4y = -(x+2)^2 + 24 \,$

$y^2-4y+4 = -(x+2)^2 + 28 \,$

$(y-2)^2 + (x+2)^2 = 28 \,$

Multiply both sides of the inequality by both numerator and denominator of both sides to get:

$(1+t^2)(1-s^2) > (1-t^2)(1+s^2) \,$

Then, expand and simplify:

$1-s^2+t^2-(ts)^2 > 1+s^2-t^2-(ts)^2 \,$

$-s^2+t^2 > s^2-t^2 \,$

$-2s^2+t^2 > -t^2 \,$

$-2s^2 > -2t^2 \,$

$s^2 < t^2 \,$

$s < t \,$