Question

Without finding the square find the difference of the following
(i) ${47}^2-{46}^2$
(ii)${52}^2-{51}^2$
(iii) ${68}^2-{67}^2$
(iv) ${73}^2-{72}^2$
(v) ${84}^2-{83}^2$
(vi) ${91}^2-{90}^2$
(vii) ${100}^2-{99}^2$
(viii) ${212}^2-{211}^2$

Answer 1

Use identity $a^2-b^2=(a+b)(a-b)$

$\left(i\right){47}^2-{46}^2=(47+46)(47-46)=\left(93\right)\left(1\right)=93$

$\left(ii\right){52}^2-{51}^2=(52+51)(52-51)=\left(103\right)\left(1\right)=103$

$\left(iii\right){68}^2-{67}^2=(68+67)(68-67)=\left(135\right)\left(1\right)=135$

$\left(iv\right){73}^2-{72}^2=(73+72)(73-72)=\left(145\right)\left(1\right)=145$

$\left(v\right){84}^2-{83}^2=(84+83)(84-83)=\left(167\right)\left(1\right)=167$

$\left(vi\right){91}^2-{90}^2=(91+90)(91-90)=\left(181\right)\left(1\right)=181$

$\left(vii\right){100}^2-{99}^2=(100+99)(100-99)=\left(199\right)\left(1\right)=199$

$\left(viii\right){212}^2-{211}^2=(212+211)(212-211)=\left(423\right)\left(1\right)=423$