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Question: Find the following squares using identities. (i) \[{(b - 7)^2}\] (ii) \({(xy + 3z)^2}\) (iii) ...

Find the following squares using identities.
(i) (b−7)2{(b - 7)^2}
(ii) (xy+3z)2{(xy + 3z)^2}
(iii) (6x2−5y)2{(6{x^2} - 5y)^2}
(iv) (23m+32n)2{(\dfrac{2}{3}m + \dfrac{3}{2}n)^2}
(v) (0.4p−0.5q)2{(0.4p - 0.5q)^2}
(vi) (2xy+5y)2{(2xy + 5y)^2}

Explanation

Solution

In the question itself it is said to use identities for finding the squares. There are algebraic identities for (a+b)2{(a + b)^2} and (a−b)2{(a - b)^2}. Using these and simplifying we can find the answers.

Useful formula:
For any a,ba,b we have these identities.
(a+b)2=a2+2ab+b2−−−(i){(a + b)^2} = {a^2} + 2ab + {b^2} - - - (i)
(a−b)2=a2−2ab+b2−−−(ii){(a - b)^2} = {a^2} - 2ab + {b^2} - - - (ii)

Complete step by step solution:
Here we are given six questions to find the square using algebraic identities.
We can do one by one.
(i) (b−7)2{(b - 7)^2}
Here we can use our second identity.
(a−b)2=a2−2ab+b2−−−(ii){(a - b)^2} = {a^2} - 2ab + {b^2} - - - (ii)
Comparing with the question we have, a=b,b=7a = b,b = 7
⇒(b−7)2=b2−2×b×7+72\Rightarrow {(b - 7)^2} = {b^2} - 2 \times b \times 7 + {7^2}
Simplifying we get,
⇒(b−7)2=b2−14b+49\Rightarrow {(b - 7)^2} = {b^2} - 14b + 49

(ii) (xy+3z)2{(xy + 3z)^2}
Here we can use our first identity.
(a+b)2=a2+2ab+b2−−−(i){(a + b)^2} = {a^2} + 2ab + {b^2} - - - (i)
Comparing with the question we have, a=xy,b=3za = xy,b = 3z
(xy+3z)2=(xy)2+2×xy×3z+(3z)2{(xy + 3z)^2} = {(xy)^2} + 2 \times xy \times 3z + {(3z)^2}
Simplifying we get,
⇒(xy+3z)2=x2y2+6xyz+9z2\Rightarrow {(xy + 3z)^2} = {x^2}{y^2} + 6xyz + 9{z^2}

(iii) (6x2−5y)2{(6{x^2} - 5y)^2}
Here we can use our second identity.
(a−b)2=a2−2ab+b2−−−(ii){(a - b)^2} = {a^2} - 2ab + {b^2} - - - (ii)
Comparing with the question we have, a=6x2,b=5ya = 6{x^2},b = 5y
(6x2−5y)2=(6x2)2−2×6x2×5y+(5y)2{(6{x^2} - 5y)^2} = {(6{x^2})^2} - 2 \times 6{x^2} \times 5{y^{}} + {(5y)^2}
Simplifying we get,
(6x2−5y)2=36x4−60x2y+25y2{(6{x^2} - 5y)^2} = 36{x^4} - 60{x^2}{y^{}} + 25{y^2}

(iv) (23m+32n)2{(\dfrac{2}{3}m + \dfrac{3}{2}n)^2}
Here we can use our first identity.
(a+b)2=a2+2ab+b2−−−(i){(a + b)^2} = {a^2} + 2ab + {b^2} - - - (i)
Comparing with the question we have, a=23m,b=32na = \dfrac{2}{3}m,b = \dfrac{3}{2}n
(23m+32n)2=(23m)2+2×23m×32n+(32n)2{(\dfrac{2}{3}m + \dfrac{3}{2}n)^2} = {(\dfrac{2}{3}m)^2} + 2 \times \dfrac{2}{3}m \times \dfrac{3}{2}n + {(\dfrac{3}{2}n)^2}
Simplifying we get,
(23m+32n)2=49m2+2mn+94n2{(\dfrac{2}{3}m + \dfrac{3}{2}n)^2} = \dfrac{4}{9}{m^2} + 2mn + \dfrac{9}{4}{n^2}

(v) (0.4p−0.5q)2{(0.4p - 0.5q)^2}
Here we can use our second identity.
(a−b)2=a2−2ab+b2−−−(ii){(a - b)^2} = {a^2} - 2ab + {b^2} - - - (ii)
Comparing with the question we have, a=0.4p,b=0.5qa = 0.4p,b = 0.5q
(0.4p−0.5q)2=(0.4p)2−2×0.4p×0.5q+(0.5q)2{(0.4p - 0.5q)^2} = {(0.4p)^2} - 2 \times 0.4p \times 0.5q + {(0.5q)^2}
Simplifying we get,
(0.4p−0.5q)2=0.16p2−0.4pq+0.25q2{(0.4p - 0.5q)^2} = 0.16{p^2} - 0.4pq + 0.25{q^2}

(vi) (2xy+5y)2{(2xy + 5y)^2}
Here we can use our first identity.
(a+b)2=a2+2ab+b2−−−(i){(a + b)^2} = {a^2} + 2ab + {b^2} - - - (i)
Comparing with the question we have, a=2xym,b=5ya = 2xym,b = 5y
(2xy+5y)2=(2xy)2+2×2xy×5y+(5y)2{(2xy + 5y)^2} = {(2xy)^2} + 2 \times 2xy \times 5y + {(5y)^2}
Simplifying we get,
(2xy+5y)2=4x2y2+20xy2+25y2{(2xy + 5y)^2} = 4{x^2}{y^2} + 20x{y^2} + 25{y^2}
⇒(2xy+5y)2=y2(4x2+20x+25)\Rightarrow {(2xy + 5y)^2} = {y^2}(4{x^2} + 20x + 25)

Additional information: These identities can be extended to any number of terms like (a+b+c)2{(a + b + c)^2} by taking two terms together at a time.

Note: We can possibly make a mistake in cases when the terms itself are squares like in question (iii). In that case, squaring the term will lead to fourth power. Also these two identities can be united. We have a−b=a+(−b)a - b = a + ( - b). So instead of the second identity we can simply replace bb by −b - b in the first identity.