Solutions To Mathematics Textbooks/Basic Mathematics/Chapter 8

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Chapter 8[edit]

Section 3[edit]

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

13[edit]

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[edit]

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[edit]

2[edit]

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 \,