Practical uses for Algebra

Algebra been used in all parts of our life activities either directly or indirectly.  If we want to find some interest which may recur with amount, we can apply progressions.  To see some different arrangements or combinations between available sources, we apply permutation and combination. To prove a generalized statement, we can apply mathematical Induction so on.

Let us see few problems of this kind.
Example Problem on Practical Uses for Algebra:

Ex 1: Peter buys a truck for dollar 120,000.  He pays half of the amount by cash. That is, dollar 60,000.  The remainingdollar 60,000, he pays in 12 annual installments of dollar 5000 each.  If the rate of interest is 12% and he pays with the installment the interest due on the unpaid amount, find the total cost of the truck.

Sol:  First installment = 5000 + 12% of 60,000

= 5000 + `12/100 xx` 60000

= $ 12,200.

Second Installment = 5000 + 12% of 55,000   = 5000 + `12 /100 xx` 55,000

= $ 11600.

Third Installment   = 5000 + `12/100 xx` 50,000

= $ 11,000

Since the common difference of the amount is $600,

Total cost of the shop = 60,000 + sum of the 12 installments.

= 60,000 + `12/2` [2` xx` 12,200 + (12 – 1) (- 600)]

= 60,000 + 6 [24,400 – 6,600]

= $ 1, 66,800.
More Example Problem on Practical Uses for Algebra:

2. A father is three times a old as his son.  In 12 years time, he will be twice as old as his son.  Find the present ages of father and son.

Solution: Let x and y be the present ages of father and his son respectively.

Given:  x = 3y `implies` x – 3 y = 0 -------- (1)

Also, x + 12 = 2 (y + 12)` implies` x + 12 = 2y + 24

`implies` x – 2y = 12 -------------------------- (2)

Therefore (1) – (2) `implies` x – 3y = 0

- x + 2y = - 12

`implies` - y = - 12 `implies` y = 12.

Therefore (1)` implies` x – 3 (12) = 0

`implies` x = 36.

Therefore the age of the father is 36 and his son’s age is 12.

Steps to Solving Quadratic Equations

In mathematics, an equation with second degree and one variable is called quadratic equation. The general equation for quadratic equation is given by,

A x2 + B x + C =0

Here x represents the variable and A, B, and C are constants, with a ≠ 0. (When a = 0, the equation is named as linear equation.)

The constant A, B, and C are expressed as quadratic coefficient, linear coefficient and the constant expression or release expression.

Steps to Solving Quadratic Equations Example 1:

5x² - x – 6 = 0, solving the factor for the given quadratic equation.

Solution:

Now, we can find the factor for the given quadratic equation

5x2 + 5x - 6x - 6 = 0

Now get the value for 5x from the primary term and 6 from secondary term.

5x (x + 1) - 6(x + 1) = 0

Now we can combine the similar term (x +1)

(5x - 6) (x + 1) = 0

To get the value for x we can associate the factor to zero

x + 1 = 0   or   5x – 6 = 0

x = - 1        or   5x = 6

x = 6 / 5

Thus, the factors  x1 and x2 are -1, 6/5.
Steps to solving quadratic equations Example 2:

3x² - 6 = -3x solving the factor for the given quadratic equation and solving the sum and product of quadratic equation.

Solution:

Fetch the -3x over: 3x² + 3x - 6 = 0

Separate 3x as 6x and -3x,

3x² + 6x - 3x - 6 = 0

Take out 3x from first two terms and 3 as common from next two terms

3x(x + 2) – 3(x + 2)  = 0

Thus, the factors are: (3x - 3) (x + 2) = 0

Modulate both expressions to zero: 3x - 3 = 0 and x + 2 = 0

3x - 3 = 0            x + 2 = 0

3x = 3                   x = - 2

x =  3/3

x = 1

So, the factors for x are  1, -2

Sum of the roots:

To find sum of roots consider the factor as x1 and x2

The sum of the roots  = x1 + x2 = (1) + (-2)

Sum of the roots = -1

Product of the roots:

To find product of roots consider the factors as x1 and x2

The product of the roots is given by x1x2 = (1)(-2)

Product of roots = -2


Quadratic Equation

ax2 +bx + c = 0


Values of

‘a’, ‘b’ and ‘c’


One Root

(x1)


Other Root

(x2)


Sum of Roots

(x1 + x2)


Product of Roots

(x1x2)

3x² +3x - 6 = 0


a = 3, b = 3, c = -6


1


-2


-1


-2

Steps to Solving Quadratic Equation Example 3:

Solving the value for x, to the given quadratic equation x2 + 5x + 4 = 0.

Solution:

Steps 1: To find the factor for the given quadratic equation, find the multiplicative value for the 4 and the sum of root value for 5

x2 + 1x + 4x + 4 = 0

Steps 2: Now obtain x as similar from the primary term and 4 as similar from last two term.

x (x +1) + 4(x + 1) = 0

Steps 3: Now we can unite similar term (x +1)

(x + 1) (x + 4) = 0

Steps 4: To obtain the value of x we can associate the factor to zero

x + 1 = 0   or   x + 4 = 0

x = - 1        or   x = -4.

Steps 5: Thus, the factors are x are -1, - 4.