Using a2-b2-a-ba-b- findi 512-492 ii 1-022-0-982 iii 1532-1472iv 12-12-7-92

(i) 512 minus; 492 = (51 + 49) (51 minus; 49) = (100) (2) = 200 (ii) (1.02)2 minus; (0.98)2 = (1.02 + 0.98) (1.02 shy; minus; 0.98) = (2) (0.04 ...

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Addition and Subtraction of Negative Numbers Using Algebra

Using a2-b2=a...

Question

Using a²− b² = (a + b) (a − b), find

(i) 51² − 49² (ii) (1.02)² − (0.98)² (iii) 153² − 147²

(iv) 12.1² − 7.9²

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Solution

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a^{2} - b^{2} = (a + b) (a - b)

(i) 153^{2} - 147^{2} (i i) {12.1}^{2} - {7.9}^{2}

Using a²-b²=(a+b)(a-b) find:

i. 51² - 49²

ii. (1.02)² - (0.98)²

Using $a^{2} - b^{2} = (a + b) (a - b)$ , find
${1.02}^{2} - {0.98}^{2}$

a^{2} - b^{2} = (a + b) (a - b)

{12.1}^{2} - {7.9}^{2}

Using the identity a^{2} - b^{2} = (a + b) (a - b)

Find the value of (1.02)^{2} - (0.98)^{2} .

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