LHS = A(B + C)
B + C = \(\left[ \begin{matrix} 1 & 2 \\ -4 & 2 \end{matrix} \right] +\left[ \begin{matrix} -7 & 6 \\ 3 & 2 \end{matrix} \right] =\left[ \begin{matrix} -6 & 8 \\ -1 & 4 \end{matrix} \right] \)
A(B + C) = \(\left[ \begin{matrix} 1 & 1 \\ -1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} -6 & 8 \\ -1 & 4 \end{matrix} \right] =\left[ \begin{matrix} -6-1 & 8+4 \\ 6-3 & -8+12 \end{matrix} \right] =\left[ \begin{matrix} -7 & 12 \\ 3 & 4 \end{matrix} \right] \) ......(1)
RHS = AB + AC
AB = \(\left[ \begin{matrix} 1 & 1 \\ -1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 2 \\ -4 & 2 \end{matrix} \right] =\left[ \begin{matrix} 1-4 & 2+2 \\ -1-12 & -2+6 \end{matrix} \right] =\left[ \begin{matrix} -3 & 4 \\ -13 & 4 \end{matrix} \right] \)
AC = \(\left[ \begin{matrix} 1 & 1 \\ -1 & 3 \end{matrix} \right] \times \left[ \begin{matrix} -7 & 6 \\ 3 & 2 \end{matrix} \right] =\left[ \begin{matrix} -7+3 & 6+2 \\ 7+9 & -6+6 \end{matrix} \right] =\left[ \begin{matrix} -4 & 8 \\ 16 & 0 \end{matrix} \right] \)
Therefore, AB + AC = \(\left[ \begin{matrix} -3 & 4 \\ -13 & 4 \end{matrix} \right] +\left[ \begin{matrix} -4 & 8 \\ 16 & 0 \end{matrix} \right] =\left[ \begin{matrix} -7 & 12 \\ 3 & 4 \end{matrix} \right] \) ....(2)
From (1) and (2), A(B + C) = AB + AC. Hence proved.

Let the height of the wall AB = x feet
As per the given data BC = (x – 7) feet
In the right triangle ABC, AC =17 ft, BC = (x – 7) feet
By Pythagoras theorem, AC² = AB² + BC²
(17)² = x²+ (x − 7)²; 289 = x² + x² − 14x + 49
x² − 7x − 120 = 0 hence, (x − 15)(x + 8) = 0 then, x = 15 (or) −8
Therefore, height of the wall AB = 15 ft (Rejecting −8 as height cannot be negative)

Because the given polynomial is a perfect square a - 16 = 0, b - 16 = 0
Therefore, a = 16, b = 16.

Therefore, \(\sqrt { 64{ x }^{ 2 }-16{ x }^{ 3 }+17{ x }^{ 2 }-2x+1 } \) = |8x² - x + 1|

\(\frac { { x }^{ 2 }+20x+36 }{ { x }^{ 2 }-3x-28 } -\frac { { x }^{ 2 }+12x+4 }{ { x }^{ 2 }-3x-28 } =\frac { \left( { x }^{ 2 }+20x+36 \right) -\left( { x }^{ 2 }+12x+4 \right) }{ { x }^{ 2 }-3x-28 } \)
= \(\frac { 8x+32 }{ { x }^{ 2 }-3x-28 } =\frac { 8\left( x+4 \right) }{ \left( x-7 \right) \left( x+4 \right) } =\frac { 8 }{ x-7 } \)

Let the present age of father be x years and the present age of son be y years
Given, x = 6y … (1)
x + 6 = 4(y + 6) … (2)
Substituting (1) in (2), 6y + 6 = 4(y + 6)
6y + 6 = 4y + 24 gives, y = 9
Therefore, son’s age = 9 years and father’s age = 54 years.

\(\frac { (k-12) }{ a } { x }^{ 2 }+\frac { 2(k-12) }{ b } x+\frac { 2 }{ c } =0\)
D² = b²- 4ac = (2(k - 12))² - 4(k - 12)(2)
= 4(k - 12)[(k - 12) - 2]
= 4(k-12)(k- 14)
The given equation will have equal roots, if D = 0
⇒ 4(k-12)(k-14) 0
k - 12 = 0 or k - 14 0
k 12, 14

9x²-9(a+b)x+(2a²+5ab+2b²)=0
Comparing this with ax² + bx + c = O.
a =9
b = -9(a + b)
c = (2a² + 5ab + 2b²)
∴ ∆=B²-4AC
⇒ 81(a+b)²-36(2a²+5ab+2b²)
⇒ 9a² + 9b² - 18ab
⇒ 9(a - b)2> 0
∴ the roots are real and given by
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 12a+6b }{ 18 } =\frac { 2a+b }{ 3 } \)
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 6a+12b }{ 18 } =\frac { a+2b }{ 3 } \)

x + y + z = 6            ....(1)
2x + 3y + 4z = 20     ...(2)
3x + 2y + 5z = 22   ....(3)

Sub. z = 3 in (5) ⇒ y - 2(3) =-4
y=2
Sub. y = 2, z = 3 in (1), we get
x+2+3=6
x=1
x= 1,y = 2, z = 3

3x² + 7x - 2 = 0 here, a = 3, b = 7, c = -2
since, α, β are the roots of the equation
\(\frac { { \alpha }^{ 2 } }{ \beta } +\frac { { \beta }^{ 2 } }{ \alpha } =\frac { { \alpha }^{ 3 }+{ \beta }^{ 3 } }{ \alpha \beta } =\frac { { \left( \alpha +\beta \right) }^{ 3 }-3\alpha \beta \left( \alpha +\beta \right) }{ \alpha \beta } =\frac { { \left( -\frac { 7 }{ 3 } \right) }^{ 3 }-3\left( -\frac { 2 }{ 3 } \right) \left( -\frac { 7 }{ 3 } \right) }{ -\frac { 7 }{ 3 } } =\frac { 469 }{ 18 } \)

x² + 7x + 10 here, a = -1, b = 7, c = 10
if α and β are roots of the equation then,
α + β = \(\frac {-b}{a} = \frac {-7}{1}\) = -7; αβ = \(\frac {c}{a} = \frac {10}{1}\) = 10
α⁴ + β⁴ = (α² + β²)² - 2α²β² = 29² - 2 x (10)² = 641 (since from (ii), α² + β² = 29)

\(\frac { x }{ { x }^{ 2 }+1 } \)
Here x² ≥ 0 for all x. Therefore, x² + 1 ≥ 0 + 1 = 1. Hence, x² + 1 ≠ 0 for any x.
Therefore, there can be no real excluded values for the given rational expression \(\frac { x }{ { x }^{ 2 }+1 } \).

The general 3 x 3 matrix is given by A = \(\left( \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right) \)  a_ij = i²j²
a₁₁ = 1² x 1² = 1 x 1 = 1; a₁₂ = 1² x 2² = 1 x 4 = 4; a₁₃ = 1² x 3² = 1 x 9 = 9
a₂₁ = 2² x 1² = 2 x 1 = 2; a₂₂ = 2² x 2² = 4 x 4 = 16; a₂₃ = 2² x 3² = 4 x 9 = 36
a₃₁ = 3² x 1² = 3 x 1 = 3; a₃₂ = 3² x 2² = 9 x 4 = 36; a₃₃ = 3² x 3² = 9 x 9 = 81
Hence the required matrix is A = \(\left( \begin{matrix} 1 & 4 & 9 \\ 4 & 16 & 36 \\ 9 & 36 & 81 \end{matrix} \right) \)

x² - 13x + k = 0 here, a = 1, b = -13, c = k
Let α, β be the roots of the equation. Then
α + β = \(\frac {-b}{a} = \frac {-(-13)}{1}\) = 13  ...... (1) also α - β = 17 ......(2)
(1) + (2) we get, 2α = 30 gives α = 15
Therefore, 15 + β = 13 (from (1)) gives β = -2
But, αβ = \(\frac {c}{a} = \frac {k}{1}\) gives 15 x (-2) = k we get, k = -30

2m² + 19m + 30 = 2m² + 4m + 15m + 30 = 2m(m + 2) + 15(m + 2)
= (m + 2)(2m + 15)
Now, equating the factors to zero we get,
(m + 2)(2m + 15) = 0
m + 2 gives, m = -2 or 2m + 15 = 0 we get, m = \(\frac {-15}{2}\)
Therefore the roots are -2, \(\frac {-15}{2}\)
Some equations which are not quadratic can be solved by reducing them to quadratic equations by suitable substitutions. Such examples are illustrated below.