Algebra Geometry Practice

Algebra is a branch of math that deals with properties of lines, curves and surfaces generally divided into two types, Algebraic and geometry in agree with the mathematical techniques. The relationship between Algebra and geometry was introduced by Descartes.

Geometry is a module of mathematics, which involves the study of shapes, line equation, angles problem, dimensions, relative position of figures etc.  The ‘Geometry’ means study of properties. A point is used to denote a position in space. such that all points lying on the line joining any two points on the surface. The algebra geometry example problems and practice problems are given below.
Example Problems for Algebra Geometry

Example problem 1:

Find the values of a and b if 3x4 + x3 + ax2 + 5x + b is exactly divisible by x + 2 and x – 1.

Let P(x) = 3x4 + x3 + ax2 + 5x + b

(x+2) and (x–1) are factors of P(x) and hence both P(–2) and P(1) = 0

P(–2) = 3(–2)4 + (–2)3 + a(–2)2 + 5(–2) + b = 30 + 4a + b

Since P(–2) = 0 we get

4a + b = –30 (1)

P(1) = 3(1)4 + (1)3 + a(1)2 + 5(1) + b = 9 + a + b

Since P(1) = 0 we get

9 + a + b = 0 or a + b = –9 (2)

(1) – (2) gives

4a + b = –30

`(a+b=-9)/(3a = 21)`

` After solving this, We get,`

` a = 7`

Substituting a = –7 in equation (2) we get

–7 + b = –9 (or) ∴ b = –2

a = –7, b = –2.

Example problem 2:

Find the remainder when x3 – 5x2 + 7x – 4 is divided by (x – 1).

Solution:

P(x) = x3 – 5x2 + 7x – 4. By using the remainder theorem when P(x) is divided by (x – 1), the remainder is P(1).

The remainder P(1) = 13 – 5(1)2 + 7(1) – 4 = –1.

Example problem 3:

If all the sides of a parallelogram touch a circle prove that the parallelogram is a rhombus.

Solution:


Given: All the sides of a parallelogram ABCD touch a circle with centre O.

To prove: ABCD is a rhombus.

Proof: Since the lengths of the tangents from an external point to a given circle are equal (Fig)

AP = AS; BP = BQ; CR = CQ; DR = DS

Adding (AP+BP)+ (CR+DR) = (AS+DS) + (BQ+CQ)

AB + CD = AD + BC (Fig)

AB + AB = AD+AD (ABCD is a parallelogram CD= AB, BC=AD)

2AB = 2AD AB= AD

But AB= CD and AD = BC ( opposite sides are equal)

AB = BC = CD = AD. Hence ABCD is a rhombus.
Practice Problems for Algebra Geometry:

Practice problem 1:

Find the quotient and the remainder using synthetic division

1.  x3 + x2 – 3x + 5 ÷ x – 1

2. 4x3 – 3x2 + 2x – 4 ÷ x – 3

Answer: 1. Remainder = 4, Quotient = x2 + 2x + 1

2. Remainder = 83, Quotient = 4x2 + 9x + 29

Practice problem 2:

In the Fig AB is the diameter of the circle, ∠BAC = 42°. Find ∠ACD

Answer: m∠ACD = 48°