PRACTICE QUESTIONS FOR a MINUS b WHOLE SQUARE

Problem 1 :

Expand (P - 2T)

Solution :

Here a  =  P and b  =  2T

(P - 2 T)2  =  P2 - 2 (P)(2T) + (2T)2

  =  P2 - 4 PT + 4T2

Problem 2 :

Expand (2P - Q)²

Solution :

Here a  =  2P and b  =  Q

(2 P - Q)2  =  (2P)2 - 2 (2P)(Q) + Q2

  =  4P2 - 4PQ + Q2

Problem 3 :

Expand (2a - 3b)²

Solution :

here a = 2a and b = 3b

(2a - 3b)2 = (2a)2 - 2 (2a)(3b) + (3b)2

  = 4a2 - 12ab + 9b2

Problem 4 :

Expand (5N - 3P)²

Solution :

here a = 5N and b  =  3P

(2a - 3b)2  =  (5N)2 - 2(5N)(3P) + (3P)2

  =  25N2 - 30NP + 9P2

Problem 5 :

Expand (9T - Q)²

Solution :

here a  =  9T and b  =  Q

(9T - Q)2  =  (9T)2 - 2 (9T)(Q) + Q2

  = 81T2 - 18 TQ + Q2

Problem 6 :

Expand (13p - 5q)²

Solution :

Here a  =  13 P and b  =  5 Q

(13P - 5Q)2  =  (13P)2 - 2 (13P)(5Q) + (5Q)2

  =  169P2 - 130 PQ + 25 Q 2

Problem 7 :

Expand (3p - 4q)²

Solution :

Here a  =  3p and b  =  4q

(3p - 4q)2  =  (3p)2 - 2(3p)(4q) + (4q)2

  =  9P2 - 24pq + 16q 2

Problem 8 :

x2 + 1/x2 = 47, then the value of x + 1/x

Solution :

Using the formula (a + b)2 = a2 + 2ab + b2

a2 + b2 (a + b)2 - 2ab

x2 + 1/x2 = (x + 1/x)2 - 2x(1/x)

47 = (x + 1/x)2 - 2

(x + 1/x)2 = 47 + 2

(x + 1/x)2 = 49

x + 1/x = √49

x + 1/x = -7 and 7

Problem 9 :

x2 + 1/x2 = 66, then the value of x - 1/x

Solution :

Using the formula (a - b)2 = a2 - 2ab + b2

a2 + b2 (a - b)2 + 2ab

x2 + 1/x2 = (x - 1/x)2 + 2x(1/x)

66 = (x - 1/x)2 + 2

(x - 1/x)2 = 66 - 2

(x - 1/x)2 = 64

x - 1/x = √64

x - 1/x = -8 and 8

Problem 10 :

Using the identity for the square of a binomial, evaluate the following 

(0.98)2

Solution :

(0.98)2 = (1 - 0.02)2

= (1 - 2/100)2

Writing the expansion of (a - b)2

= 12 - 2(1)(2/100) + (2/100)2

= 1 - (4/100) + (4/10000)

= 1 - 0.04 + 0.0004

= 1.0004 - 0.04

= 0.9604

Problem 11 :

Find the square of (3x/4 - 4y/5)

Solution :

(3x/4 - 4y/5)2

Here a = 3x/4 and b = 4y/5

= (3x/4)2 - 2(3x/4)(4y/5) + (4y/5)2

= (9x2/16) - (24xy/20) + (16y2/25)

= (9x2/16) - (6xy/5) + (16y2/25)

Problem 12 :

236 x 236 - 2 (236)(86) + 86 x 86

Solution :

236 x 236 - 2 (236)(86) + 86 x 86

= (236 - 86)2

= 1502

= 22500

Problem 13 :

1.06 x 1.06 - 2 (1.06)(0.06) + 0.06 x 0.06

Solution :

1.06 x 1.06 - 2 (1.06)(0.06) + 0.06 x 0.06

= (1.06 - 0.06)2

= 12

= 1

Problem 14  :

287 x 287 + 269 x 269 - 2 x 287 x 269

Solution :

= 287 x 287 + 269 x 269 - 2 x 287 x 269

= 2872 + 2692 - 2 (287)(269)

= (287 - 269)2

= 182

= 324

Problem 14 :

If a - b = 3 and a2 + b2 = 29, find the value of ab

Solution :

Given that, a - b = 3 and a2 + b2 = 29

a2 + b2 = (a - b)2 + 2ab

Applying these values in the formula, we get

29 = 32 + 2ab

29 - 9 = 2ab

2ab = 20

ab = 20/2

ab = 10

So, the value of ab is 10.

Problem 15 :

Expand and simplify (x + 3)2 + (x - 4)2

Solution :

= (x + 3)2 + (x - 4)2

= x+ 2x(3) + 32 + x- 2x(4) + 42

= x+ 6x + 9 + x- 8x + 16

= 2x- 2x + 25

Problem 15 :

Rationalize the denominator 1/(√9 - √8)

Solution :

1/(√9 - √8)

To rationalize the denominator, we should multiply both numerator and denominator by the conjugate of the denominator.

= [1/(√9 - √8)] [(√9 + √8)/(√9 - √8)]

= [(√9 + √8)/(√9 + √8)(√9 - √8)]

(√9 + √8)/(√92 - √82)

(√9 + √8)/(9 - 8)

(√9 + √8)/1

(= √9 + √8)

Related topics

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Digital SAT Math Problems and Solutions (Part - 190)

    Jun 19, 25 08:35 PM

    digitalsatmath257.png
    Digital SAT Math Problems and Solutions (Part - 190)

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 189)

    Jun 19, 25 07:30 PM

    Digital SAT Math Problems and Solutions (Part - 189)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 188)

    Jun 17, 25 02:26 AM

    digitalsatmath254.png
    Digital SAT Math Problems and Solutions (Part - 188)

    Read More