VEDIC MATHS-92

 

VEDIC MATHS

                           By OMKAR TENDOLKAR

Hello friends,

                      This is  post number 92 from the series of "Vedic maths" blogs. Here in this blog we will learn about "Algebraic Calculations-II"

Reference:

We had already learn about "Algebraic Calculations-I" in our previous blog. If you have missed my last blog then please visit "VEDIC MATHS-91".

Algebraic Calculations:

Here in this post we will solve algebraic calculations by maths sutra.

Examples :

1. Solve (2x - 3)^2 - (3x - 2)(x + 5).

Step 1: 

We solve this using the vertically and crosswise method. First, we multiply vertically the terms that have x. So we have

(2x - 3)

(2x - 3)

2x × 2x = 4(x)^2

And

(3x - 2)

(x + 5)

Here, multiplying vertically we get 3x × x = 3(x)^2.

So mentally, we do it like this: 4(x)^2 - 3(x)^2 = (x)^2.

This is the first part of our answer.

Step 2:

Next, we multiply crosswise.

So, as the first part, we have

(2x - 3)

(2x - 3)

So we get from the first product -2x × 3 and -2x × 3 = -6x + -6x = -12x.

And from the second product, we get 5 × (3x) - 2 × x.

(3x - 2)

(x + 5)

So we have 15x - 2x = 13x.

So combining the first part with the second part we have -12x + -13x = -25x.

Step 3:

And in the final step, we add the independent terms, +9 and +10 and get 19.

So our answer is (2x - 3)^2 – (3x - 2)(x + 5) = (x)^2 - 25x + 19.

Answer:

 (2x - 3)^2 – (3x - 2)(x + 5) = (x)^2 - 25x + 19.

2. Solve (5x - 2)^2 - (2x + 1)^2.

Step 1: 

Using the same method, we first multiply vertically the terms that have x.

So we have

(5x +2)

(5x + 2)

So 5x × 5x = 25(x)^2.

And we have

(2x + 1)

(2x + 1)

So 2x × 2x = 4(x)^2.

Now we’ll take the terms containing (x)^2 : 25(x)^2 - 4(x)^2

(from the second product) = 21(x)^2.

Step 2:

Now in this step, we multiply crosswise. So we have

(5x + 2)

(5x + 2)

Here we multiply 5x × 2 twice = 10x + 10x = 20x. Again we have

(2x + 1)

(2x + 1)

Here we cross-multiply 2x × 1, both ways, and get 2x both times.

We add 2x and 2x and get 4x.

Altogether, this becomes 20x - 4x = 16x.

Step 3:

And in our final step, we have the independent digits 22 - 12 = 4 - 1 = 3.

So our answer is (5x + 2)^2 - (2x + 1)^2 = 21(x)^2 + 16x + 3.

Answer:

(5x + 2)^2 - (2x + 1)^2 = 21(x)^2 + 16x + 3.

Solve the following example.

1. (x – 4)^2 + (x – 1)^2

Ans : (x – 4)^2 + (x – 1)^2 = 2(x)^2 - 10x – 17.

2. (3x – 6)^2 + (x – 5)^2

Ans : (3x – 6)^2 + (x – 5)^2 =  10(x)^2 - 46x – 61.

3. (3x – 6)^2 + (2x – 5)^2

Ans :  (3x – 6)^2 + (2x – 5)^2= 13(x)^2 - 56x – 61.

The simplicity of this method can be vouched from examples given above.

You may try following examples.

1. (x – 4)^2 + (x – 3)^2

2. (2x – 4)^2 + (2x – 5)^2

3. (2x – 5)^2 + (3x – 1)^2

4. (4x – 4)^2 + (4x – 1)^2

5. (8x – 4)^2 + (9x – 1)^2

In next blog we will discuss about "Algebraic Calculations-III".     

Are you excited for this?...

Then, please wait for it.

I will post my new blog in next week.

We will meet very soon through our next  blog. Till that stay connected, stay healthy and stay safe.

Thanks

for giving your valuable time.

Good day😊