D=b²-4ac=0
Here a=1+m²,b=2mc,c=(c²-a²)=0
(2mc)²-4(1+m²)(c²-a²)=0
\(\Rightarrow 4m^{ 2 }c^{ 2 }-4(1+m^{ 2 })(c^{ 2 }-a^{ 2 })\)
\(\Rightarrow m^{ 2 }c^{ 2 }-4(1+m^{ 2 })(c^{ 2 }-a^{ 2 })=0\)
\(\Rightarrow m^{ 2 }c^{ 2 }-(c^{ 2 }-a^{ 2 }+m^{ 2 }c^{ 2 }-m^{ 2 }a^{ 2 })=0\)
\(\Rightarrow m^{ 2 }c^{ 2 }-(c^{ 2 }+a^{ 2 }+m^{ 2 }c^{ 2 }+m^{ 2 }a^{ 2 })=0\)
\(\Rightarrow -c^{ 2 }+a^{ 2 }+m^{ 2 }a^{ 2 }=0\)
c²=a²(1+m²)
Hence proved.

Let the numbers be x, x + 2
(x)²+(x+2)²=340
x²+x²+4+4x=340
\(\Rightarrow 2x^{ 2 }+4x-3369=0\)
On dividing by 2, we get
x²+2x-168=0
\(\Rightarrow (x+14)(x-12)=0\)
\(\Rightarrow x=12\)
The numbers are, 12, (12 + 2) i.e., 12, 14

13 cm, 15 cm

Let shortest side be x unit! and other side be units Hypotenuse = z units 
ATQ x+y+z = 5x
\(\Rightarrow \) y+z=4x
\(\Rightarrow \) z=4x-y
Also, area of the rectangle = 15x 
\(\Rightarrow \) \(\frac { 1 }{ 2 } x.y=15x\Rightarrow y=30\) ...(i)
Using Pythagoras Theorem, we get 
z²=x²+y²
\(\Rightarrow \) (4x-y)² = x²+y²
\(\Rightarrow \)  (4x - 30)² =x²+(30)² [using (i)]
\(\Rightarrow 16x^{ 2 }-240x+900=x^{ 2 }+900\)
\(\Rightarrow 15x^{ 2 }-240x=0\Rightarrow 15x(x-16)=0\)
\(\Rightarrow \) x=0 (rejecting) or x=16
\(\therefore x=16\)
\(\therefore \) length of the shortest side = 16 units
length of other side = 30 units
length of hypotenuse z = 4 x 16 - 30
= 34 units 

Consider the equation \(\Rightarrow \frac { x-2 }{ x-3 } +\frac { x-4 }{ x-5 } =\frac { 10 }{ 3 } \)
\(\Rightarrow 1+\frac { 1 }{ x-3 } +1+\frac { 1 }{ x-5 } =\frac { 10 }{ 3 } \)
\(\Rightarrow \frac { 1 }{ x-3 } +\frac { 1 }{ x-5 } =\frac { 10 }{ 3 } \)
\(\Rightarrow \frac { x-5+x-3 }{ (x-3)(x-5) } =\frac { 4 }{ 3 } \)
\(\Rightarrow \frac { 2x-8 }{ x^{ 2 }-8x+15 } =\frac { 4 }{ 3 } \)
\(\Rightarrow 4x^{ 2 }-32x+60=6x-24\)
\(\Rightarrow 4x^{ 2 }-38x+84=0\)
\(\Rightarrow 2x^{ 2 }-19x+42=0\)
\(\Rightarrow 2x^{ 2 }-12x-7x+42=0\)
\(\Rightarrow 2x^{ 2 }-12x-7x+42=0\)
\(\Rightarrow 2x(x-6)-7(x-6)=0\)
\(\Rightarrow (2x-7)(x-6)=0\)
\(\Rightarrow \) Either 2x-7 =0 x-6=0
\(\Rightarrow x=\frac { 7 }{ 2 } ,6\)

Let original duration of the flight be x hours.
 Distance = 2800 km 
\(\therefore \) Usual method = \(\frac { 2800 }{ x } km/h\)
When time = \(\left( x+\frac { 1 }{ 2 } \right) \) hrs
The new speed = \(\frac { 2800 }{ x+\frac { 1 }{ 2 } } =\frac { 5600 }{ 2x+1 } \)
ATQ \(\frac { 2800 }{ x } -\frac { 5600 }{ 2x+1 } =100\quad \Rightarrow \frac { 2800(2x+1)-5600x }{ (2x+1)x } =100\)
\(\Rightarrow 2800=100(2x^{ 2 }+x)\Rightarrow 2x^{ 2 }+x-28=0\)
\(\Rightarrow 2x^{ 2 }+8x-7x-28=0\quad \Rightarrow 2x(x+4)-7(x+4)=0\)
\(\Rightarrow (x+4)(2x-7)=0\quad \Rightarrow x=-4\) (rejected) or x=\(\frac { 7 }{ 2 } =3\frac { 1 }{ 2 } \)
\(\therefore \) Original duration = \(3\frac { 1 }{ 9 } \) hours

Side of larger square =8cm; Side of smaller square =5cm.

x²+5x-(a²+a-6)=0
D=(5)²-4x1x[-(a²+a-6)]
=25+4a²+4a-24
=4a²+4a+1=(2a+1)²
\(\therefore x=\frac { -b\pm \sqrt { D } }{ 2a } =\frac { -5\pm \sqrt { (2a+1) } ^{ 2 } }{ 2\times 1 } \)
\(=\frac { -5\pm (2a+1) }{ 2 } =\frac { -4+2a }{ 2 } ,\frac { -6-2a }{ 2 } \)
=-2+a,-3-a

Given quadratic equation is 2x²-6x+a=0
\(\alpha ,\beta \) are roots of the equation
\(\alpha +\beta =\frac { -b }{ a } \Rightarrow \alpha +\beta =-\frac { -6 }{ 2 } \Rightarrow \alpha +\beta =3\Rightarrow \alpha =3-\beta \)............... (i)
Also \(2\alpha +5\beta =12\)
\(\Rightarrow 2\left( 3-\beta \right) +5=12\)
\(\Rightarrow 6-2\beta +5\beta =12\Rightarrow 3\beta =6\Rightarrow \beta =2\)
where \(\beta =2\), eq.(i) becomes \(\alpha =3-2=1\)
Now,product of roots = \(\frac { c }{ a } \)
\(\Rightarrow \alpha .\beta =\frac { a }{ 2 } \Rightarrow 1\times 2=\frac { a }{ 2 } \Rightarrow a=4\)