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)

Distance d = t²- 0.75t,
Given that d = 11.25 = t²- 0.75t.
\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(=\frac { (+0.75)\pm \sqrt { { (-0.75) }^{ 2 }-4\times 1\times -11.25 } }{ 2\times 1 } \)
\(=\frac { +0.75\pm \sqrt { 0.5625+45 } }{ 2 } \)
\(=\frac { +0.75\pm \sqrt { 45.5625 } }{ 2 } \)
\(=\frac { +0.75\pm 6.75 }{ 2 } \)
\(=\frac { 7.50 }{ 2 } or\frac { -6 }{ 2 } \)
= 3.75 or-3 It is not possible
∴ t = 3.75 s.

Compare the coefficients of the given equation with the standard form ax² + bx + c = 0
a = pq, b = -(p + q)², c = (p + q)²
x = \({-b \pm \sqrt{b^2-4ac} \over 2a}\)
substituting the values of a, b and c in the formula we get,
x = \(\frac { -\left[ -{ \left( p+q \right) }^{ 2 } \right] \pm \sqrt { { \left[ -{ \left( p+q \right) }^{ 2 } \right] }^{ 2 }-4\left( pq \right) { \left( p+q \right) }^{ 2 } } }{ 2pq } \)
= \(\frac { { \left( p+q \right) }^{ 2 }\pm \sqrt { { \left( p+q \right) }^{ 4 }-4\left( pq \right) { \left( p+q \right) }^{ 2 } } }{ 2pq } \)
= \(\frac { { \left( p+q \right) }^{ 2 }\pm \sqrt { { \left( p+q \right) }^{ 2 }\left[ { \left( p+q \right) }^{ 2 }-4pq \right] } }{ 2pq } \)
= \(\frac { { \left( p+q \right) }^{ 2 }\pm \sqrt { { \left( p+q \right) }^{ 2 }\left( { p }^{ 2 }+{ q }^{ 2 }+2pq-4pq \right) } }{ 2pq } \)
= \(\frac { { \left( p+q \right) }^{ 2 }\pm \sqrt { { \left( p+q \right) }^{ 2 }{ \left( p+q \right) }^{ 2 } } }{ 2pq } \)
= \(\frac { { \left( p+q \right) }^{ 2 }\pm \left( p+q \right) \left( p-q \right) }{ 2pq } =\frac { \left( p+q \right) \left\{ \left( p+q \right) \pm \left( p-q \right) \right\} }{ 2pq } \)
Therefore, x = \(\frac { p+q }{ 2pq } \times 2p,\frac { p+q }{ 2pq } \times 2q\) we get, x = \(\frac { p+q }{ q } ,\frac { p+q }{ p } \)

\(\frac { { x }^{ 2 }+6x+8 }{ { x }^{ 8 }+8 } \) -q(x)\(\frac { 3 }{ { x }^{ 2 }-2x+4 } \)
\(q(x)=\frac { { x }^{ 2 }+6x+8 }{ { x }^{ 3 }+8 } -\frac { 3 }{ { x }^{ 2 }-2x+4 } \)
\(=\frac { \left( { x }^{ 2 }+6x+8 \right) ({ x }^{ 3 }-2x+4)-3\left( { x }^{ 3 }+8 \right) }{ \left( { x }^{ 3 }+8 \right) \left( { x }^{ 2 }-2x+4 \right) } \)
\(=\frac { { x }^{ 4 }+6{ x }^{ 2 }+8{ x }^{ 2 }-2{ x }^{ 3 }-12{ x }^{ 2 }-16x+4{ x }^{ 2 }+24x+32-3({ x }^{ 3 }+8) }{ (x+8)({ x }^{ 2 }-2x+4) } \)
\(=\frac { { x }^{ 4 }+{ 4x }^{ 3 }+8x+32-3({ x }^{ 3 }+8) }{ ({ x }^{ 3 }+8)({ x }^{ 2 }-2x+4) } \)
\(=\frac { { x }^{ 3 }(x+4)+8(x+4)-3({ x }^{ 3 }+8) }{ ({ x }^{ 3 }+8)({ x }^{ 2 }-2x+4) } \)
\(=\frac { (x+4)({ x }^{ 3 }+8)-3({ x }^{ 3 }+8) }{ ({ x }^{ 3 }+8)({ x }^{ 2 }-2x+4) } \)
\(=\frac { ({ x }^{ 3 }+8)(x+4-3) }{ ({ x }^{ 3 }+8)({ x }^{ 2 }-2x+4) } \)
\(=\frac { x+1 }{ ({ x }^{ 2 }+2x+4) } \)

p(x)=x²-5x-14
\(({ x }^{ 2 }-5x-14)\div q(x)=\frac { x-7 }{ x+2 } \)
\(({ x }^{ 2 }-5x-14)\times \frac { 1 }{ q(x) } =\frac { x-7 }{ x+2 } \)
\(\frac { 1 }{ q(x) } =\frac { x-7 }{ x+2 } \times \frac { 1 }{ { x }^{ 2 }-5x-14 } \)
\(\frac{1}{q(x)}=\frac{\not x - 7}{x+2} \times \frac{1}{(\not x-7)(x+2)}=\frac{1}{(x+2)^{2}}\)
∴ q(x)=(x+2)²=x²+4x+4

x⁴ + 3x³ - x - 3, x³ + x² - 5x + 3
Let us divide the highest degree polynomial by least degree polynomial and
Let f(x) = x⁴ + 3x³ - x - 3 
g(x) = x³ + x² - 5x + 3
Now, dividing f (x) by g (x).

Since '3 'is not the DMSOR of g (x),let us divide g (x) by x² + 2x - 3.

Since, the Remainder is zero, the GCD is x² + 2x - 3.

Let the 100t digit be 'x', 10's be 'y' and Unit digit be 'z'.
The three digit number is 100x + 10y + z.
Now given, x + y + z = 11 .       ..(1)
100z +10y + x = 5 (100x + 10y + z) + 46
Simplifying
499x + 40y - 95z = - 46          .........(2)
x + 2y = z
x + 2y - z = 0    ....(3)

Substituting the value x = 1 in(4)
2(1) + 3y = 11
3y = 11 - 2 = 9
\(y=\frac{9}{3}=3\)
Substituting x = 1, y = 3 in (1)
1 + 3 + z = 11
z = 11 - 4 = 7
x = 1, y = 3, z = 7
The original three digit number is 137
i.e., 100(1) + 10(3) + 1(7) = 100+ 30 + 7 =137

Let the present ages of Vani, her father and her grand father be x, y, z respectively
Given
\(\frac { x+y+z }{ 3 } =53\Rightarrow x+y+z=159\quad \quad ...(1)\)
\(\frac{z}{2}+\frac{y}{3}+\frac{x}{4}=65\)
6z + 4y + 3x = 780          ...(2)
(z - 4) = 4(x - 4)
4x - z = 12       ..........(3)
Consider (1) and (3


Substituting x = 24 in (4)
5(24) + y = 171
y = 171 - 120 = 51
Substituting x = 24, y = 51 in (1)
24 + 51 + z = 159
z = 159 - 75 = 84
Present age of Vani = 24 years
Present age of her father = 51 years
Present age of her grandfather = 84 years.

Let, x + 2y - z = 5... (1)  
x - y + z = -2....(2)  
-5x - 4y + z = -11... (3)


Here we arrive at an identity 0 = 0
Hence the system has an infinite number of solutions.

\(A=\left[ \begin{matrix} 3 & 0 \\ 4 & 5 \end{matrix} \right] ,B=\left[ \begin{matrix} 6 & 3 \\ 8 & 5 \end{matrix} \right] ,C=\left[ \begin{matrix} 3 & 6 \\ 1 & 1 \end{matrix} \right] \)
CD - AB = 0 ⇒ CD = AB
\(AB=\left[ \begin{matrix} 3 & 0 \\ 4 & 5 \end{matrix} \right] \left[ \begin{matrix} 6 & 3 \\ 8 & 5 \end{matrix} \right] =\left[ \begin{matrix} (18+0) & (9+0) \\ (24+40) & (12+25) \end{matrix} \right] \)
\(CD=\left[ \begin{matrix} 18 & 9 \\ 64 & 37 \end{matrix} \right] \)
\(Let\quad D=\left[ \begin{matrix} x & y \\ z & w \end{matrix} \right] \)
\(\left[ \begin{matrix} 3 & 6 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x & y \\ z & w \end{matrix} \right] =\left[ \begin{matrix} 18 & 9 \\ 64 & 37 \end{matrix} \right] \)
\(\left[ \begin{matrix} 3x+6z & 3y+6w \\ x+z & y+w \end{matrix} \right] =\left[ \begin{matrix} 18 & 9 \\ 64 & 37 \end{matrix} \right] \)
3x + 6z = 18      ...(1)
x + z = 64    ...(2)

Sub. x=122 in(2)
122+z=64
z=64-122=-58
3y+6w=9  ....(3)
 y+w=37   ....(4)

Sub. w = -34 in (4)
y-34 = 37
y = 37 + 34 = 71
∴ Solutions: x = 122
y = 71
z = -58
w = -34
\(\therefore D=\left[ \begin{matrix} 122 & 71 \\ -58 & -34 \end{matrix} \right] \)