Exercise 3.1: Algebraic Expressions (Squares and Expressions Involving Squares)
1. Find the square with the help of the formulae:
1) $2a + 3b$
Solution $(2a + 3b)^2$
$= (2a)^2 + 2 \cdot (2a) \cdot (3b) + (3b)^2$
$= 4a^2 + 12ab + 9b^2$
Answer:$4a^2 + 12ab + 9b^2$
2) $x^2 + \frac{2}{y^2}$
Solution: $\left(x^2 + \frac{2}{y^2}\right)^2$
$= (x^2)^2 + 2 \cdot (x^2) \cdot \left(\frac{2}{y^2}\right) + \left(\frac{2}{y^2}\right)^2$
$= x^4 + \frac{4x^2}{y^2} + \frac{4}{y^4}$
Answer:$x^4 + \frac{4x^2}{y^2} + \frac{4}{y^4}$
3) $4y – 5x$
Solution: $(4y – 5x)^2$
$= (4y)^2 – 2 \cdot (4y) \cdot (5x) + (5x)^2$
$= 16y^2 – 40xy + 25x^2$
Answer:$16y^2 – 40xy + 25x^2$
4) $5x^2 – y$
Solution: $(5x^2 – y)^2$
$= (5x^2)^2 – 2 \cdot (5x^2) \cdot y + y^2$
$= 25x^4 – 10x^2y + y^2$
Answer:$25x^4 – 10x^2y + y^2$
5) $3b – 5c – 2a$
Solution: $(3b – 5c – 2a)^2$
$= \{(3b – 5c) – 2a\}^2$
$= (3b – 5c)^2 – 2 \cdot (3b – 5c) \cdot (2a) + (2a)^2$
$= (3b)^2 – 2 \cdot (3b) \cdot (5c) + (5c)^2 – 4a(3b – 5c) + 4a^2$
$= 9b^2 – 30bc + 25c^2 – 12ab + 20ca + 4a^2$
$= 4a^2 + 9b^2 + 25c^2 – 12ab – 30bc + 20ca$
Answer:$4a^2 + 9b^2 + 25c^2 – 12ab – 30bc + 20ca$
6) $ax – by – cz$
Solution: $(ax – by – cz)^2$
$= \{ax – (by + cz)\}^2$
$= (ax)^2 – 2 \cdot (ax) \cdot (by + cz) + (by + cz)^2$
$= a^2x^2 – 2abxy – 2acxz + b^2y^2 + 2bcyz + c^2z^2$
$= a^2x^2 + b^2y^2 + c^2z^2 – 2abxy + 2bcyz – 2acxz$
Answer:$a^2x^2 + b^2y^2 + c^2z^2 – 2abxy + 2bcyz – 2acxz$
7) $2a + 3x – 2y – 5z$
Solution: $(2a + 3x – 2y – 5z)^2$
$= \{(2a + 3x) – (2y + 5z)\}^2$
$= (2a + 3x)^2 – 2(2a + 3x)(2y + 5z) + (2y + 5z)^2$
$= (4a^2 + 12ax + 9x^2) – 2(4ay + 10az + 6xy + 15xz) + (4y^2 + 20yz + 25z^2)$
$= 4a^2 + 9x^2 + 4y^2 + 25z^2 + 12ax – 8ay – 20az – 12xy – 30xz + 20yz$
Answer:$4a^2 + 9x^2 + 4y^2 + 25z^2 + 12ax – 8ay – 20az – 12xy – 30xz + 20yz$
8) $1007$
Solution: $(1007)^2$
$= (1000 + 7)^2$
$= (1000)^2 + 2 \cdot 1000 \cdot 7 + 7^2$
$= 1000000 + 14000 + 49$
$= 1014049$
Answer:$1014049$
2. Simplify:
1) $(7p + 3q – 5r)^2 – 2(7p + 3q – 5r)(8p – 4q – 5r) + (8p – 4q – 5r)^2$
Solution: Let, $7p + 3q – 5r = a$ and $8p – 4q – 5r = b$
Given expression $= a^2 – 2ab + b^2$
$= (a – b)^2$
$= \{(7p + 3q – 5r) – (8p – 4q – 5r)\}^2$
$= (7p + 3q – 5r – 8p + 4q + 5r)^2$
$= (-p + 7q)^2$
$= (7q – p)^2$
$= (7q)^2 – 2(7q)(p) + p^2$
$= 49q^2 – 14pq + p^2$
Answer:$p^2 – 14pq + 49q^2$
2) $(2m + 3n – p)^2 + (2m – 3n + p)^2 – 2(2m + 3n – p)(2m – 3n + p)$
Solution: Let, $2m + 3n – p = a$ and $2m – 3n + p = b$
Given expression $= a^2 + b^2 – 2ab$
$= (a – b)^2$
$= \{(2m + 3n – p) – (2m – 3n + p)\}^2$
$= (2m + 3n – p – 2m + 3n – p)^2$
$= (6n – 2p)^2$
$= (6n)^2 – 2(6n)(2p) + (2p)^2$
$= 36n^2 – 24np + 4p^2$
Answer:$36n^2 – 24np + 4p^2$
3) $6.35 \times 6.35 + 2 \times 6.35 \times 3.65 + 3.65 \times 3.65$
Solution: Let, $a = 6.35$ and $b = 3.65$
Given expression $= a^2 + 2ab + b^2$
$= (a + b)^2$
$= (6.35 + 3.65)^2$
$= (10)^2$
$= 100$
Answer:$100$
4) $\frac{2345 \times 2345 – 759 \times 759}{2345 – 759}$
Solution: Let, $a = 2345$ and $b = 759$
Given expression $= \frac{a^2 – b^2}{a – b}$
$= \frac{(a + b)(a – b)}{a – b}$
$= a + b$
$= 2345 + 759$
$= 3104$
Answer:$3104$
3. If $a – b = 4$ and $ab = 60$, what is the value of $a + b$?
Solution: We know that, $(a + b)^2 = (a – b)^2 + 4ab$
$= (4)^2 + 4 \times 60$
$= 16 + 240$
$= 256$
$\implies a + b = \pm \sqrt{256} = \pm 16$
Answer:$\pm 16$
4. If $a + b = 9m$ and $ab = 18m^2$, what is the value of $a – b$?
Solution:
We know that, $(a – b)^2 = (a + b)^2 – 4ab$
$= (9m)^2 – 4(18m^2)$
$= 81m^2 – 72m^2$
$= 9m^2$
$\implies a – b = \pm \sqrt{9m^2} = \pm 3m$
Answer:$\pm 3m$
5. If $x – \frac{1}{x} = 4$, prove that $x^4 + \frac{1}{x^4} = 322$.
Solution:
Left Hand Side $= x^4 + \frac{1}{x^4}$
$= (x^2)^2 + \left(\frac{1}{x^2}\right)^2$
$= \left(x^2 + \frac{1}{x^2}\right)^2 – 2 \cdot x^2 \cdot \frac{1}{x^2}$
$= \left\{\left(x – \frac{1}{x}\right)^2 + 2 \cdot x \cdot \frac{1}{x}\right\}^2 – 2$
$= \{(4)^2 + 2\}^2 – 2$
$= (16 + 2)^2 – 2$
$= (18)^2 – 2 = 324 – 2 = 322 = \text{Right Hand Side}$ (Proved)
6. If $2x + \frac{2}{x} = 3$, what is the value of $x^2 + \frac{1}{x^2}$?
Solution:
Given, $2\left(x + \frac{1}{x}\right) = 3$
$\implies x + \frac{1}{x} = \frac{3}{2}$
Given expression $= x^2 + \frac{1}{x^2}$
$= \left(x + \frac{1}{x}\right)^2 – 2 \cdot x \cdot \frac{1}{x}$
$= \left(\frac{3}{2}\right)^2 – 2$
$= \frac{9}{4} – 2$
$= \frac{9 – 8}{4}$
$= \frac{1}{4}$
Answer:$\frac{1}{4}$
7. If $a + \frac{1}{a} = 2$, show that $a^2 + \frac{1}{a^2} = a^4 + \frac{1}{a^4}$
Solution:
Left Hand Side $= a^2 + \frac{1}{a^2}$
$= \left(a + \frac{1}{a}\right)^2 – 2$
$= (2)^2 – 2$
$= 4 – 2$
$= 2$
Right Hand Side $= a^4 + \frac{1}{a^4}$
$= \left(a^2 + \frac{1}{a^2}\right)^2 – 2$
$= (2)^2 – 2$
$= 4 – 2$
$= 2$
Therefore, Left Hand Side $=$ Right Hand Side (Shown)
8. If $a + b = \sqrt{7}$ and $a – b = \sqrt{5}$, prove that $8ab(a^2 + b^2) = 24$
Solution:
Left Hand Side $= 8ab(a^2 + b^2)$
$= 4ab \cdot 2(a^2 + b^2)$
$= \{(a + b)^2 – (a – b)^2\} \cdot \{(a + b)^2 + (a – b)^2\}$
$= \{(\sqrt{7})^2 – (\sqrt{5})^2\} \cdot \{(\sqrt{7})^2 + (\sqrt{5})^2\}$
$= (7 – 5)(7 + 5)$
$= 2 \times 12$
$= 24 = \text{Right Hand Side}$ (Proved)
9. If $a + b + c = 9$ and $ab + bc + ca = 31$, what is the value of $a^2 + b^2 + c^2$?
Solution:
We know that, $a^2 + b^2 + c^2 = (a + b + c)^2 – 2(ab + bc + ca)$
$= (9)^2 – 2(31)$
$= 81 – 62$
$= 19$
Answer:$19$
10. If $a^2 + b^2 + c^2 = 9$ and $ab + bc + ca = 8$, what is the value of $(a + b + c)^2$?
Solution:
We know that, $(a + b + c)^2 = (a^2 + b^2 + c^2) + 2(ab + bc + ca)$
$= 9 + 2(8)$
$= 9 + 16$
$= 25$
Answer:$25$
11. If $a + b + c = 6$ and $a^2 + b^2 + c^2 = 14$, what is the value of $(a – b)^2 + (b – c)^2 + (c – a)^2$?
Solution:
We know that, $2(ab + bc + ca) = (a + b + c)^2 – (a^2 + b^2 + c^2)$
$= 6^2 – 14$
$= 36 – 14$
$= 22$
Given expression $= (a – b)^2 + (b – c)^2 + (c – a)^2$
$= (a^2 – 2ab + b^2) + (b^2 – 2bc + c^2) + (c^2 – 2ca + a^2)$
$= 2(a^2 + b^2 + c^2) – 2(ab + bc + ca)$
$= 2(14) – 22$
$= 28 – 22$
$= 6$
Answer:$6$
12. If $x = 3, y = 4$, and $z = 5$, what is the value of $9x^2 + 16y^2 + 4z^2 – 24xy – 16yz + 12zx$?
Solution:
Given expression $= (3x)^2 + (-4y)^2 + (2z)^2 + 2(3x)(-4y) + 2(-4y)(2z) + 2(2z)(3x)$
$= (3x – 4y + 2z)^2$
Substituting $x = 3, y = 4, z = 5$:
$= (3(3) – 4(4) + 2(5))^2$
$= (9 – 16 + 10)^2$
$= (3)^2$
$= 9$
Answer:$9$
13. Express $(a + 2b)(3a + 2c)$ as the difference of two squares.
Solution:
Let $x = a + 2b$ and $y = 3a + 2c$
We know that, $xy = \left(\frac{x + y}{2}\right)^2 – \left(\frac{x – y}{2}\right)^2$
Here,
$x + y = (a + 2b) + (3a + 2c) = 4a + 2b + 2c$
$\implies \frac{x + y}{2} = \frac{4a + 2b + 2c}{2} = 2a + b + c$
and
$x – y = (a + 2b) – (3a + 2c) = -2a + 2b – 2c$
$\implies \frac{x – y}{2} = \frac{-2a + 2b – 2c}{2} = b – a – c$
Therefore,
$(a + 2b)(3a + 2c) = (2a + b + c)^2 – (b – a – c)^2$
Answer:$(2a + b + c)^2 – (b – a – c)^2$
14. Express $x^2 + 10x + 24$ as the difference of two squares.
Solution:
$x^2 + 10x + 24$
$= x^2 + 10x + 25 – 1$
$= (x)^2 + 2 \cdot x \cdot 5 + (5)^2 – (1)^2$
$= (x + 5)^2 – (1)^2$
Answer:$(x + 5)^2 – 1^2$
If $a^4 + a^2b^2 + b^4 = 8$ and $a^2 + ab + b^2 = 4$, find the values of,
1) $a^2 + b^2$
2) $ab$
Solution:
Given, $a^4 + a^2b^2 + b^4 = 8$
$\implies (a^2 + b^2)^2 – a^2b^2 = 8$
$\implies (a^2 + ab + b^2)(a^2 – ab + b^2) = 8$
$\implies 4 \times (a^2 – ab + b^2) = 8$
$\implies a^2 – ab + b^2 = 2$ …….. (i)
Given, $a^2 + ab + b^2 = 4$ …….. (ii)
1) Adding equations (i) and (ii):
$2(a^2 + b^2) = 6 \implies a^2 + b^2 = 3$
Answer (a):$3$
2) Subtracting equation (i) from (ii):
$2ab = 2 \implies ab = 1$
Answer (b):$1$