Algebraic Identities Worksheet 1
1. Find the following by using identity - I:
(i) \( (2x+5)^2 \)
Solution
\[ \begin{align*} \color{green} (a+b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = 2x \, , \, b = 5 \\ \\ & = (2x+5)^2 \\ & = (2x)^2 + 2 \times 2x \times 5 + (5)^2 \\ & = \color{red}4x^2 + 20x + 25 \end{align*} \]
(ii) \( (8x + 3y)^2 \)
Solution
\[ \begin{align*} \color{green} (a + b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = 8x \, , \, b = 3y \\ \\ & = (8x + 3y)^2 \\ & = (8x)^2 + 2 \times 8x \times 3y + (3y)^2 \\ & = \color{red}64x^2 + 48xy + 9y^2 \end{align*} \]
(iii) \( \left(\frac{3}{5}a + \frac{2}{3}b\right)^2 \)
Solution
\[ \begin{align*} \color{green} (a + b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = \frac{3}{5}a \, , \, b = \frac{2}{3}b \\ \\ & = \left(\frac{3}{5}a + \frac{2}{3}b\right)^2 \\ & = \left(\frac{3}{5}a\right)^2 + 2 \times \frac{\cancel3}{5}a \times \frac{2}{\cancel3}b + \left(\frac{2}{3}b\right)^2 \\ & = \color{red}\frac{9}{25}a^2 + \frac{4}{5}ab + \frac{4}{9}b^2 \end{align*} \]
(iv) \( (7pq + 4ab)^2 \)
Solution
\[ \begin{align*} \color{green} (a + b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = 7pq \, , \, b = 4ab \\ \\ & = (7pq + 4ab)^2 \\ & = (7pq)^2 + 2 \times 7pq \times 4ab + (4ab)^2 \\ & = \color{red}49p^2q^2 + 56pqab + 16a^2b^2 \end{align*} \]
(v) \( (0.2x + 1.5y)^2 \)
Solution
\[ \begin{align*} \color{green} (a + b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = 0.2x \, , \, b = 1.5y \\ \\ & = (0.2x + 1.5y)^2 \\ & = (0.2x)^2 + 2 \times 0.2x \times 1.5y + (1.5y)^2 \\ & = \color{red}0.04x^2 + 0.6xy + 2.25y^2 \end{align*} \]
(vi) \( (2m^2 + 3n^2)^2 \)
Solution
\[ \begin{align*} \color{green} (a + b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = 2m^2 \, , \, b = 3n^2 \\ \\ & = (2m^2 + 3n^2)^2 \\ & = (2m^2)^2 + 2 \times 2m^2 \times 3n^2 + (3n^2)^2 \\ & = \color{red}4m^4 + 12m^2n^2 + 9n^4 \end{align*} \]
2. Evaluate the following by using identity - I:
(i) \( (101)^2 \)
Solution
\[ \begin{align*} (101)^2 & = (100 + 1)^2 \\ \\ \color{green} (a+b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = 100 \, , \, b = 1 \\ \\ & = (100 + 1)^2 \\ & = (100)^2 + 2 \times 100 \times 1 + (1)^2 \\ & = 10000 + 200 + 1 \\ (101)^2 & = \color{red} 10201 \\ \end{align*} \]
(ii) \( (52)^2 \)
Solution
\[ \begin{align*} (52)^2 & = (50 + 2)^2 \\ \\ \color{green} (a+b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = 50 \, , \, b = 2 \\ \\ & = (50 + 2)^2 \\ & = (50)^2 + 2 \times 50 \times 2 + (2)^2 \\ & = 2500 + 200 + 4 \\ (52)^2 & = \color{red} 2704 \\ \end{align*} \]
(iii) \( (8.1)^2 \)
Solution
\[ \begin{align*} (8.1)^2 & = (8 + 0.1)^2 \\ \\ \color{green} (a+b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = 8 \, , \, b = 0.1 \\ \\ & = (8 + 0.1)^2 \\ & = (8)^2 + 2 \times 8 \times 0.1 + (0.1)^2 \\ & = 64 + 1.6 + 0.01 \\ (8.1)^2 & = \color{red} 65.61 \\ \end{align*} \]
(iv) \( (203)^2 \)
Solution
\[ \begin{align*} (203)^2 & = (200 + 3)^2 \\ \\ \color{green} (a+b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = 200 \, , \, b = 3 \\ \\ & = (200 + 3)^2 \\ & = (200)^2 + 2 \times 200 \times 3 + (3)^2 \\ & = 40000 + 1200 + 9 \\ (203)^2 & = \color{red} 41209 \\ \end{align*} \]
(v) \( (410)^2 \)
Solution
\[ \begin{align*} (410)^2 & = (400 + 10)^2 \\ \\ \color{green} (a+b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = 400 \, , \, b = 10 \\ \\ & = (400 + 10)^2 \\ & = (400)^2 + 2 \times 400 \times 10 + (10)^2 \\ & = 160000 + 8000 + 100 \\ (410)^2 & = \color{red} 168100 \\ \end{align*} \]
(vi) \( (10.2)^2 \)
Solution
\[ \begin{align*} (10.2)^2 & = (10 + 0.2)^2 \\ \\ \color{green} (a+b)^2 &= \color{green} a^2 + 2ab + b^2 \\ & a = 10 \, , \, b = 0.2 \\ \\ & = (10 + 0.2)^2 \\ & = (10)^2 + 2 \times 10 \times 0.2 + (0.2)^2 \\ & = 100 + 4 + 0.04 \\ (10.2)^2 & = \color{red} 104.04 \\ \end{align*} \]