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:

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

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

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

$-2s^2>-2t^2$

$s^2<t^2$

$s<t$