\(x^{ 2 }-26x+160=0\)

(i) No
(ii) Yes
(iii) No
(iv) No
(v) Yes

\(Real \ and \ distinct \ \left [\sqrt{6} , - {\sqrt {6}\over 3}\right ]\)

\(\left [{5\over 2} , \ 5\right]\)

\(\left [{5\over 4}, {4\over 5}\right ]\)

\(\left [3\pm \sqrt {19}\over 5\right]\)

Let age of the son be x years
Father's agebe (x)²
Also x² + 5x = 66 
\(\Rightarrow x^{ 2 }+5x-66=0\)
\(\Rightarrow (x+11)(x-6)=0\)
\(\Rightarrow x=-11,x=6\)
\(\therefore \quad \)  Age of son =6 years
and age of Father = (6)²=36 years

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

Kx²+1-2(k-41)x+x²=0
\(\Rightarrow kx^{ 2 }+x^{ 2 }-2(k-1)x+1=0\)
\(\Rightarrow (k+1)x^{ 2 }-2(k-1)x+1=0\)
Here,a=k +1 ,b=-2(k-1),c=1
D=b²-4ac
=[-2(k-1)²--4(k+1)(1)
=4(k-1)²-4(k+1)
Roots are equal
D=0
\(\Rightarrow 4(k-1)^{ 2 }-4(k+1)=0\)
\(\Rightarrow 4(k-1)^{ 2 }-(k+1)=0\)
\(\Rightarrow (k-1)^{ 2 }-k-1=0\)
\(\Rightarrow k^{ 2 }+1-2k-k-1=0\)
\(\Rightarrow k^{ 2 }-3k=0\)
\(\Rightarrow k(k-3)=0\)
k=0 or k-3=0 \(\Rightarrow k=3\)
Non zero value of k =3
after putting the value of k=3 in equation we get
3x3²+1-2(3-1)x+x² =0
\(\Rightarrow 3x^{ 2 }+1-2(2)x+x^{ 2 }=0\)
\(\Rightarrow 3x^{ 2 }+1-4x+x^{ 2 }=0\)
\(\Rightarrow 4x^{ 2 }-4x+1=0\)
\(\Rightarrow 4x^{ 2 }-2x-2x+1=0\)
\(\Rightarrow 2x(2x-1)-1(2x-1)=0\)
\(\Rightarrow (2x-1)\quad (2x-1)=0\)
2x-1=0 or 2x-1=0
2x-1 or 2x=1
\(x=\frac { 1 }{ 2 } \) or \(x=\frac { 1 }{ 2 } \)
Roots are \(x=\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \)

The equation has real roots if \(D\ge0\)
\((4)^2-4p\ge0\) 
i.e., \(16-4p\ge0 \ or \ 16\ge4p \ or \ p\le4\)

If the equation  x²+kx+64=0 has real roots,then D \(\ge 0\)
\(\Rightarrow k^{ 2 }-4\times 64\ge 0\quad \Rightarrow k^{ 2 }\ge 256\Rightarrow k^{ 2 }\ge (16)^{ 2 }\)
\(\Rightarrow k\ge 16\)   \(\left[ \therefore k>0 \right] \) ...(i)
If the equation x^{2 - 8x+k=0  has real roots then D _{\(\ge \) 0} \(\Rightarrow 64-4k\ge 0\Rightarrow 4k\le 64\) _...(ii)}
From (i) and (ii) we get k=16

Here given quadratic equation is 2x²+7x+5=0
a and b are roots a+b=\(\frac{-7}{2}\) ...(i)
and a.b=\(\frac{5}{2}\) ...(ii)
Now, quadratic equation whose roots are 2a+3 and 2b+3 is
x²-[2a+3+2b+3]x+(2a+3)(2b+3)=0
\(\Rightarrow\) x²-[2(a+b)+6]x+(4ab+6(a+b)+9]=0
\(\Rightarrow { x }^{ 2 }-\left[ 2\left( \frac { -7 }{ 2 } \right) +6 \right] x+\left[ 4\times \frac { 5 }{ 2 } +6\times \left( \frac { -7 }{ 2 } \right) +9 \right] =0\)  [using (i) and (ii)]
\(\Rightarrow\) x²+x-2=0

Let Price of a book Rs x : Total cost =Rs 300
Number of books = \(\frac { 300 }{ x } \)
If price of a book = Rs (x-5) then number of books = \(\frac { 300 }{ x-5 } \)
ATQ \(\frac { 300 }{ x-5 } -\frac { 300 }{ x } =5\Rightarrow \frac { 300-300(x-5) }{ (x-5)x } =5\)
\(\Rightarrow 1500=(5x^{ 2 }-5x)\Rightarrow x^{ 2 }-5x-300=0\)
\(\Rightarrow (x-20)(x+15)=0\Rightarrow x=20,15\) (rejected)
Hence Original list price of a book is Rs 20

Let digit at unit's place = a and digit at ten's place = y
Number = 10y+x
A.T.Q xy=18 \(\Rightarrow y=\frac { 18 }{ x } \) ....(i)
and 10y+x-63=10x+y \(\Rightarrow 9y-9x-63=0\)
\(\Rightarrow y-x-7=0\Rightarrow \frac { 18 }{ x } -x-7=0\)[Using eq(i)]
\(\Rightarrow 18-x^{ 2 }-7x=0\Rightarrow x^{ 2 }+7x-18=0\) 
\(\Rightarrow (x+9)(x-2)=0\Rightarrow x=-9,x=2\)
When x=2 ,y =  \(\frac { 18 }{ 2 } =9\)
Number = 92

16,13