Exercise 4.3 Practice
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Overview
This page provides comprehensive Ch 4: Algebraic Identities – Exercise 4.3 Practice. Practice squaring larger numbers, factoring quadratic and trinomial squares using advanced algebraic identities, and verifying algebraic equality relations with step-by-step solutions.
Trinomial Squares $(a + b + c)^2$ & Advanced Factorisation
Q1: Evaluate Squares Using Identities
Find the following squares using one of the identities. Determine which of these identities will make these calculations easier.
(i) $117^2$ (ii) $78^2$ (iii) $198^2$
(iv) $214^2$ (v) $1104^2$ (vi) $1120^2$
(i) $117^2$ (ii) $78^2$ (iii) $198^2$
(iv) $214^2$ (v) $1104^2$ (vi) $1120^2$
(i) $117^2$:
Write $117 = 120 - 3$. Apply $(a - b)^2 = a^2 - 2ab + b^2$:
$$117^2 = (120 - 3)^2 = 120^2 - 2(120)(3) + 3^2$$
$$= 14400 - 720 + 9 = \mathbf{13689}$$
Write $117 = 120 - 3$. Apply $(a - b)^2 = a^2 - 2ab + b^2$:
$$117^2 = (120 - 3)^2 = 120^2 - 2(120)(3) + 3^2$$
$$= 14400 - 720 + 9 = \mathbf{13689}$$
(ii) $78^2$:
Write $78 = 80 - 2$. Apply $(a - b)^2 = a^2 - 2ab + b^2$:
$$78^2 = (80 - 2)^2 = 80^2 - 2(80)(2) + 2^2$$
$$= 6400 - 320 + 4 = \mathbf{6084}$$
Write $78 = 80 - 2$. Apply $(a - b)^2 = a^2 - 2ab + b^2$:
$$78^2 = (80 - 2)^2 = 80^2 - 2(80)(2) + 2^2$$
$$= 6400 - 320 + 4 = \mathbf{6084}$$
(iii) $198^2$:
Write $198 = 200 - 2$. Apply $(a - b)^2 = a^2 - 2ab + b^2$:
$$198^2 = (200 - 2)^2 = 200^2 - 2(200)(2) + 2^2$$
$$= 40000 - 800 + 4 = \mathbf{39204}$$
Write $198 = 200 - 2$. Apply $(a - b)^2 = a^2 - 2ab + b^2$:
$$198^2 = (200 - 2)^2 = 200^2 - 2(200)(2) + 2^2$$
$$= 40000 - 800 + 4 = \mathbf{39204}$$
(iv) $214^2$:
Write $214 = 200 + 14$. Apply $(a + b)^2 = a^2 + 2ab + b^2$:
$$214^2 = (200 + 14)^2 = 200^2 + 2(200)(14) + 14^2$$
$$= 40000 + 5600 + 196 = \mathbf{45796}$$
Write $214 = 200 + 14$. Apply $(a + b)^2 = a^2 + 2ab + b^2$:
$$214^2 = (200 + 14)^2 = 200^2 + 2(200)(14) + 14^2$$
$$= 40000 + 5600 + 196 = \mathbf{45796}$$
(v) $1104^2$:
Write $1104 = 1100 + 4$. Apply $(a + b)^2$:
$$1104^2 = (1100 + 4)^2 = 1100^2 + 2(1100)(4) + 4^2$$
$$= 1210000 + 8800 + 16 = \mathbf{1218816}$$
Write $1104 = 1100 + 4$. Apply $(a + b)^2$:
$$1104^2 = (1100 + 4)^2 = 1100^2 + 2(1100)(4) + 4^2$$
$$= 1210000 + 8800 + 16 = \mathbf{1218816}$$
(vi) $1120^2$:
Write $1120 = 1100 + 20$. Apply $(a + b)^2$:
$$1120^2 = (1100 + 20)^2 = 1100^2 + 2(1100)(20) + 20^2$$
$$= 1210000 + 44000 + 400 = \mathbf{1254400}$$
Write $1120 = 1100 + 20$. Apply $(a + b)^2$:
$$1120^2 = (1100 + 20)^2 = 1100^2 + 2(1100)(20) + 20^2$$
$$= 1210000 + 44000 + 400 = \mathbf{1254400}$$
(i) 13689 (ii) 6084 (iii) 39204 (iv) 45796 (v) 1218816 (vi) 1254400
Q2: Factor Using Identities
Factor using suitable identities:
(i) $16y^2 - 24y + 9$ (ii) $\dfrac{9}{64}s^2 + \dfrac{3}{4}st + t^2$
(iii) $\dfrac{m^2}{9} + \dfrac{2}{3}mk + k^2 - \dfrac{n^2}{4}$ (iv) $p^2 - 8 + \dfrac{16}{p^2}$
(v) $9a^2 + 4b^2 + c^2 + 12ab - 4bc - 6ca$
(i) $16y^2 - 24y + 9$ (ii) $\dfrac{9}{64}s^2 + \dfrac{3}{4}st + t^2$
(iii) $\dfrac{m^2}{9} + \dfrac{2}{3}mk + k^2 - \dfrac{n^2}{4}$ (iv) $p^2 - 8 + \dfrac{16}{p^2}$
(v) $9a^2 + 4b^2 + c^2 + 12ab - 4bc - 6ca$
(i) $16y^2 - 24y + 9$:
Write as perfect squares: $(4y)^2 - 2(4y)(3) + 3^2$.
Applying $(a - b)^2 = a^2 - 2ab + b^2$:
$$16y^2 - 24y + 9 = \mathbf{(4y - 3)^2}$$
Write as perfect squares: $(4y)^2 - 2(4y)(3) + 3^2$.
Applying $(a - b)^2 = a^2 - 2ab + b^2$:
$$16y^2 - 24y + 9 = \mathbf{(4y - 3)^2}$$
(ii) $\dfrac{9}{64}s^2 + \dfrac{3}{4}st + t^2$:
Write as perfect squares: $\left(\frac{3}{8}s\right)^2 + 2\left(\frac{3}{8}s\right)(t) + t^2$.
Applying $(a + b)^2 = a^2 + 2ab + b^2$:
$$\dfrac{9}{64}s^2 + \dfrac{3}{4}st + t^2 = \mathbf{\left(\frac{3}{8}s + t\right)^2}$$
Write as perfect squares: $\left(\frac{3}{8}s\right)^2 + 2\left(\frac{3}{8}s\right)(t) + t^2$.
Applying $(a + b)^2 = a^2 + 2ab + b^2$:
$$\dfrac{9}{64}s^2 + \dfrac{3}{4}st + t^2 = \mathbf{\left(\frac{3}{8}s + t\right)^2}$$
(iii) $\dfrac{m^2}{9} + \dfrac{2}{3}mk + k^2 - \dfrac{n^2}{4}$:
Group the first three terms as a perfect square:
$$\left(\frac{m^2}{9} + \frac{2}{3}mk + k^2\right) - \frac{n^2}{4} = \left(\frac{m}{3} + k\right)^2 - \left(\frac{n}{2}\right)^2$$
Use difference of squares $X^2 - Y^2 = (X + Y)(X - Y)$:
$$= \mathbf{\left(\frac{m}{3} + k + \frac{n}{2}\right)\left(\frac{m}{3} + k - \frac{n}{2}\right)}$$
Group the first three terms as a perfect square:
$$\left(\frac{m^2}{9} + \frac{2}{3}mk + k^2\right) - \frac{n^2}{4} = \left(\frac{m}{3} + k\right)^2 - \left(\frac{n}{2}\right)^2$$
Use difference of squares $X^2 - Y^2 = (X + Y)(X - Y)$:
$$= \mathbf{\left(\frac{m}{3} + k + \frac{n}{2}\right)\left(\frac{m}{3} + k - \frac{n}{2}\right)}$$
(iv) $p^2 - 8 + \dfrac{16}{p^2}$:
Write as $p^2 - 2(p)\left(\frac{4}{p}\right) + \left(\frac{4}{p}\right)^2$.
Applying $(a - b)^2 = a^2 - 2ab + b^2$:
$$= \mathbf{\left(p - \frac{4}{p}\right)^2}$$
Write as $p^2 - 2(p)\left(\frac{4}{p}\right) + \left(\frac{4}{p}\right)^2$.
Applying $(a - b)^2 = a^2 - 2ab + b^2$:
$$= \mathbf{\left(p - \frac{4}{p}\right)^2}$$
(v) $9a^2 + 4b^2 + c^2 + 12ab - 4bc - 6ca$:
Write as: $(3a)^2 + (2b)^2 + (-c)^2 + 2(3a)(2b) + 2(2b)(-c) + 2(-c)(3a)$.
Applying $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$:
$$= \mathbf{(3a + 2b - c)^2}$$
Write as: $(3a)^2 + (2b)^2 + (-c)^2 + 2(3a)(2b) + 2(2b)(-c) + 2(-c)(3a)$.
Applying $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$:
$$= \mathbf{(3a + 2b - c)^2}$$
(i) $(4y - 3)^2$ (ii) $\left(\frac{3}{8}s + t\right)^2$ (iii) $\left(\frac{m}{3} + k + \frac{n}{2}\right)\left(\frac{m}{3} + k - \frac{n}{2}\right)$ (iv) $\left(p - \frac{4}{p}\right)^2$ (v) $(3a + 2b - c)^2$
Q3: Trinomial Expansion
Expand the following using the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$:
(i) $(p + 3q + 7r)^2$ (ii) $(3x - 2y + 4z)^2$
(i) $(p + 3q + 7r)^2$ (ii) $(3x - 2y + 4z)^2$
(i) $(p + 3q + 7r)^2$:
Here, $a = p$, $b = 3q$, $c = 7r$:
$$(p + 3q + 7r)^2 = p^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)$$
$$= \mathbf{p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp}$$
Here, $a = p$, $b = 3q$, $c = 7r$:
$$(p + 3q + 7r)^2 = p^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)$$
$$= \mathbf{p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp}$$
(ii) $(3x - 2y + 4z)^2$:
Here, $a = 3x$, $b = -2y$, $c = 4z$:
$$(3x - 2y + 4z)^2 = (3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x)$$
$$= \mathbf{9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx}$$
Here, $a = 3x$, $b = -2y$, $c = 4z$:
$$(3x - 2y + 4z)^2 = (3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x)$$
$$= \mathbf{9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx}$$
(i) $p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp$
(ii) $9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx$
(ii) $9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx$
Q4: Identify Verification
Is this an identity?
$$(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2$$
$$(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2$$
To check if the equation is an identity, expand the Left-Hand Side (LHS):
• $(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca$
• $(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca$
• $(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc + 2ca$
• $(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca$
• $(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca$
• $(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc + 2ca$
Add the three expansions together:
$$\text{LHS} = (a^2 + b^2 + c^2 + 2ab - 2bc - 2ca) + (a^2 + b^2 + c^2 - 2ab - 2bc + 2ca) + (a^2 + b^2 + c^2 - 2ab + 2bc + 2ca)$$
$$\text{LHS} = 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc + 2ca$$
$$\text{LHS} = (a^2 + b^2 + c^2 + 2ab - 2bc - 2ca) + (a^2 + b^2 + c^2 - 2ab - 2bc + 2ca) + (a^2 + b^2 + c^2 - 2ab + 2bc + 2ca)$$
$$\text{LHS} = 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc + 2ca$$
Since the LHS simplifies to $3a^2 + 3b^2 + 3c^2 - 2ab - 2bc + 2ca$, it is **not equal** to the Right-Hand Side (RHS) $2a^2 + 2b^2 + 2c^2$ for all values of $a, b, c$.
Therefore, the statement is **not an identity**.
Therefore, the statement is **not an identity**.
No, this is not an identity.