Solutions To Mathematics Textbooks/Basic Mathematics/Chapter 8

Chapter 8

Section 3

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

13

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

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

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

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

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

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

15

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

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

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

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

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

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

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

Section 4

2

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:

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

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

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

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

$s^{2}<t^{2}\,$

$s<t\,$