Using the identity , expand the following: (i) (ii) (iii) (iv) (v) (vi)

Understand the Question The identity for the square of a binomial is: To expand any given expression: 1. Identify the first term as and the second term as . 2. Substitute and into . 3. Simplify each term (square the coefficients and variables, and multiply the middle terms). Part (i) Expand Step 1 · Apply the Identity Here, and . Using Answer > (i) Part (ii) Expand Step 1 · Apply the Identity Here, and . Using Answer > (ii) Part (iii) Expand Step 1 · Apply the Identity Here, and .Using Answer > (iii) Part (iv) Expand Step 1 · Apply the Identity Here, and .Using Answer > (iv) Part (v) Expand Step 1 · Apply the Identity Here, and .Using Answer > (v) Part (vi) Expand Step 1 · Apply the Identity Here, and .Using Answer > (vi) Common Mistakes Missing the Middle Term: Writing by forgetting the middle term . Incomplete Squaring: Forgetting to square the numerical coefficients (e.g., writing instead of , or instead of ). Fraction Multiplication Errors: Incorrectly canceling terms in the middle product when fractions are involved.

Exploring Algebraic Identities | Exercise 4.1

Question 1

Using the identity $(a + b)^2 = a^2 + 2ab + b^2$ , expand the following:

(i) $(7x + 4y)^2$

(ii) $\left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2$

(iii) $(2.5p + 1.5q)^2$

(iv) $\left(\dfrac{3}{4}s + 8t\right)^2$

(v) $\left(x + \dfrac{1}{2y}\right)^2$

(vi) $\left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2$

Solution

Understand the Question

The identity for the square of a binomial is: $(a + b)^2 = a^2 + 2ab + b^2$
To expand any given expression:
1. Identify the first term as $a$ and the second term as $b$ .
2. Substitute $a$ and $b$ into $a^2 + 2ab + b^2$ .
3. Simplify each term (square the coefficients and variables, and multiply the middle terms).

(i) Expand $(7x + 4y)^2$

Step 1 · Apply the Identity

Here, $a = 7x$ and $b = 4y$ .

Diagram 5 Using $(a + b)^2 = a^2 + 2ab + b^2$

\begin{aligned} (7x + 4y)^2 &= (7x)^2 + 2(7x)(4y) + (4y)^2 \\[0.6em] &= 49x^2 + 56xy + 16y^2 \end{aligned}

Answer

(i) $49x^2 + 56xy + 16y^2$

(ii) Expand $\left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2$

Step 1 · Apply the Identity

Here, $a = \dfrac{7}{5}x$ and $b = \dfrac{3}{2}y$ .

Diagram 6 Using $(a + b)^2 = a^2 + 2ab + b^2$

\begin{aligned} \left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2 &= \left(\dfrac{7}{5}x\right)^2 + 2\left(\dfrac{7}{5}x\right)\left(\dfrac{3}{2}y\right) + \left(\dfrac{3}{2}y\right)^2 \\[0.6em] &= \dfrac{49}{25}x^2 + \dfrac{21}{5}xy + \dfrac{9}{4}y^2 \end{aligned}

Answer

(ii) $\dfrac{49}{25}x^2 + \dfrac{21}{5}xy + \dfrac{9}{4}y^2$

(iii) Expand $(2.5p + 1.5q)^2$

Step 1 · Apply the Identity

Here, $a = 2.5p$ and $b = 1.5q$ .Using $(a + b)^2 = a^2 + 2ab + b^2$

\begin{aligned} (2.5p + 1.5q)^2 &= (2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2 \\[0.6em] &= 6.25p^2 + 7.5pq + 2.25q^2 \end{aligned}

Answer

(iii) $6.25p^2 + 7.5pq + 2.25q^2$

(iv) Expand $\left(\dfrac{3}{4}s + 8t\right)^2$

Step 1 · Apply the Identity

Here, $a = \dfrac{3}{4}s$ and $b = 8t$ .Using $(a + b)^2 = a^2 + 2ab + b^2$

\begin{aligned} \left(\dfrac{3}{4}s + 8t\right)^2 &= \left(\dfrac{3}{4}s\right)^2 + 2\left(\dfrac{3}{4}s\right)(8t) + (8t)^2 \\[0.6em] &= \dfrac{9}{16}s^2 + 12st + 64t^2 \end{aligned}

Answer

(iv) $\dfrac{9}{16}s^2 + 12st + 64t^2$

(v) Expand $\left(x + \dfrac{1}{2y}\right)^2$

Step 1 · Apply the Identity

Here, $a = x$ and $b = \dfrac{1}{2y}$ .Using $(a + b)^2 = a^2 + 2ab + b^2$

\begin{aligned} \left(x + \dfrac{1}{2y}\right)^2 &= (x)^2 + 2(x)\left(\dfrac{1}{2y}\right) + \left(\dfrac{1}{2y}\right)^2 \\[0.6em] &= x^2 + \dfrac{x}{y} + \dfrac{1}{4y^2} \end{aligned}

Answer

(v) $x^2 + \dfrac{x}{y} + \dfrac{1}{4y^2}$

(vi) Expand $\left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2$

Step 1 · Apply the Identity

Here, $a = \dfrac{1}{x}$ and $b = \dfrac{1}{y}$ .Using $(a + b)^2 = a^2 + 2ab + b^2$

\begin{aligned} \left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2 &= \left(\dfrac{1}{x}\right)^2 + 2\left(\dfrac{1}{x}\right)\left(\dfrac{1}{y}\right) + \left(\dfrac{1}{y}\right)^2 \\[0.6em] &= \dfrac{1}{x^2} + \dfrac{2}{xy} + \dfrac{1}{y^2} \end{aligned}

Answer

(vi) $\dfrac{1}{x^2} + \dfrac{2}{xy} + \dfrac{1}{y^2}$

Common Mistakes

Missing the Middle Term: Writing $(a + b)^2 = a^2 + b^2$ by forgetting the middle term $2ab$ .
Incomplete Squaring: Forgetting to square the numerical coefficients (e.g., writing $(7x)^2 = 7x^2$ instead of $49x^2$ , or $(2y)^2 = 2y^2$ instead of $4y^2$ ).
Fraction Multiplication Errors: Incorrectly canceling terms in the middle product $2ab$ when fractions are involved.

Question 1

More questions in Exercise 4.1