Question

Solve using suitable identity.

(x + 3z)²​

Answer 1

Answer:

${x}^{2} + {9z}^{2} + 6xz$

Step-by-step explanation:

We Have :-

$( \: x + 3 {z} \: )^{2}$

To Find :-

Its value using suitable identity

Identity Used :-

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

Solution :-

$( \: x + 3 {z} \: )^{2} \\ \\ using \: identity \\ \\ {x}^{2} + {9z}^{2} + 2 \times x \times 3z \\ \\ {x}^{2} + {9z}^{2} + 6xz$

Answer 2

${\large{\frak{\pmb{Question}}}}$

• Solve the following by using suitable identity = (x+3z)²

${\large{\frak{\pmb{Using \; identity}}}}$

• (a+b)² = a² + 2ab + b²

${\large{\frak{\pmb{Solution}}}}$

• (x + 3z)² = x² + 6xz + 3z²

${\large{\frak{\pmb{Full \; Solution}}}}$

→ (x+3z)²

→ (a+b)² = a² + 2ab + b²

${\large{\frak{\pmb{Here,}}}}$

★ a is x

★ b is 3z

${\large{\frak{\pmb{Putting \; the \; values}}}}$

→ x² +2(x)(3z) + 3z²

→ x² + 6xz + 3z²

${\large{\frak{\pmb{Additional \; information}}}}$

Algebraic identities –

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A+B)^{2} \: = \: = A^{2} \: + \: 2AB \: + B^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A-B)^{2} \: = \: = A^{2} \: - \: 2AB \: + B^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto A^{2} \: - B^{2} \: = \: (A+B) \: (A-B)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A+B)^{2} \: = (A-B)^{2} \: +4AB}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A-B)^{2} \: = (A+B)^{2} \: -4AB}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A+B)^{3} \: = A^{3} + \: 3AB \: (A+B) \:+ B^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto (A-B)^{3} \: = A^{3} - \: 3AB \: (A-B) \: + B^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto A^{3} \: + B^{3} = \: (A+B) (A^{2} - AB + B^{2}}}}$

Factorised identities -

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{2}\; =\;a^{2}\:+\:b^{2}\:+\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\:b)^{2}\;=\;a^{2}\:+\:b^{2}\:-\:2ab}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\:b\:+\:c)^{2}\;=\;a^{2}\:+\:b^{2}\:+\:c^{2}\:+\:2ab\:+\:2bc\:+\:2ac}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:+\;b)^{3}\;=\;a^{3}\:+\:b^{3}\:+\:3ab(a\:+\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;(a\:-\;b)^{3}\;=\;a^{3}\:-\:b^{3}\:-\:3ab(a\:-\:b)}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)^{2} \: = \: a^{2} + 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a-b)^{2} \: = a^{2} - 2ab + b^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto \;\;(a+b)(a-b) \: = \: a^{2} - b^{2}}\end{gathered}$