Question

Using the formula for squaring a binomial evaluate the following
$(69)2$
​

Answer 1

$\tt{ {(69)}^{2} can \: be \: written \: as :}$

$\tt{ {(60+9)}^{2}}$

If we carefully observe ; the new value which we have got i.e. the binomial is in the form of $\tt{ {(a+b)}^{2}}$

So , we will use the same identity to solve the given problem.

$\tt{ {(a+b)}^{2} = {a}^{2} + 2ab + {b}^{2}}$

After putting up the values of a and b as:

a = 60

b = 9

We get :

$\tt{ {(60+9)}^{2} = {(60)}^{2} + 2 \times 60 \times 9+ {(9)}^{2}}$

$\longrightarrow{\tt{ {(60+9)}^{2} = 3600 + 1080+ 81}}$

$\large{\boxed{\implies{\tt{ {(69)}^{2} = 4761}}}}$

Verification :

$\tt{ \sqrt{4761} \implies 69}$

Answer 2

$\tt (69) ^{2} \: can \: be \: expressed \: as \: (70 - 1) ^{2}$

$\tt Solving \: the \: above \: using \: the \: formula \: (a - b) ^{2}$

As, in the above Question

$\bold a = 70 \\ \bold b = 1$

Thus, solving on the Following Identity

$\rm (a - b {)}^{2} = {a}^{2} - 2ab - + {b}^{2}$

$(70 {)}^{2} - 2 \times 70 \times 1 + (1 {)}^{2}$

$\red{ \implies} \tt 4900 - 140 + 1$

$\red{ \implies} \tt4901 - 140$

$\red{ \implies} \tt 4761$

Important Identities

$\tt(a + b{)}^{0} = 1$

$\tt(a + b {)}^{1} = a + b$

$\tt(a + b {)}^{2} = {a}^{2} + 2ab + {b}^{2}$

$\tt(a - b {)}^{2} = {a}^{2} - 2ab + {b}^{2}$

$\tt {a}^{2} - {b}^{2} = (a - b)(a +b )$

$\tt{a}^{2} + {b}^{2} = (a + b {)}^{2} - 2ab$

$\tt{a}^{2} + {b}^{2} = (a - b {)}^{2} + 2ab$

$\tt(a + b) ^{3} = {a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3}$

$\tt(a - b {)}^{3} = {a}^{3} - 3 {a}^{2} b + 3a {b}^{2} - {b}^{3}$

$\tt {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )$

$\tt{a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )$

$\tt(a + b + c) ^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca$