Question

Simplify the following using the identities:

$\left(i\right)\frac{{58}^2-{42}^2}{16}\left(ii\right)178\times 178-22\times 22\left(iii\right)\frac{198\times 198-102\times 102}{96}\left(iv\right)1.73\times 1.73-0.27\times 0.27\left(v\right)\frac{8.63\times 8.63-1.37\times 1.37}{0.726}$

Answer 1

$\left(i\right)\frac{{58}^2-{42}^2}{16}Let58=a,42=b\therefore \frac{a^2-b^2}{16}=\frac{(a+b)(a-b)}{16}=\frac{(58+42)(58-42)}{16}=\frac{100\times 16}{16}=100\left(ii\right)178\times 178-22\times 22=(178{)}^2-(22{)}^2=(178+22)(178-22)\{a^2-b^2-(a+b\left)\right(a-b\left)\right\}=200\times 156=31200\left(iii\right)\frac{198\times 198-102\times 102}{96}=\frac{(198{)}^2-(102{)}^2}{96}=\frac{(198+102)(198-102)}{96}\{a^2-b^2=(a+b\left)\right(a-b\left)\right\}=\frac{300\times 96}{96}=300\left(iv\right)1.73\times 1.73-0.27\times 0.27=(1.73{)}^2-(0.27{)}^2=(1.73+0.27)(1.73-0.27)\{a^2-b^2=(a+b\left)\right(a-b\left)\right\}=2.00\times 1.46=2.9200=2.92\left(v\right)\frac{8.63\times 8.63-1.37\times 1.37}{0.726}=\frac{(8.63{)}^2-(1.37{)}^2}{0.726}=\frac{(8.63+1.37)(8.63-1.37)}{0.726}\{a^2-b^2=(a+b\left)\right(a-b\left)\right\}=\frac{10.00\times 7.26}{0.726}=10.00\times 10=100$