Putting x = -\(\frac{1}{2}\). in 3x² + 2kx - 3 = 0
\(3\left( -\frac { 1 }{ 2 } \right) ^{ 2 }+2k\left( -\frac { 1 }{ 2 } \right) -3=0\)
⇒ \(\frac{3}{4}\) - k- 3 = 0
⇒ k = \(\frac{3}{4}\)- 3
⇒ k = \(\frac{3-12}{4}\)
⇒ k = \(\frac{-9}{4}\)

Find D then use formula \(\frac { -b\pm \sqrt { D } }{ 20 } \) .\(\frac { c }{ b } ,\frac { -b }{ a } \)

-1

Since 1 is a root of equation ay²+ay+3=0,
\(\therefore\) a(1)²+a(1)+3=0
\(\Rightarrow\) 2a+3=0 \(\Rightarrow\) a=\(\frac {-3}{2}\)
Also, 1 is a root y²+y+b=0
\(\therefore\) (1)²+(1)+b=0
\(\Rightarrow\) 2+b=0 \(\Rightarrow\) b=-2
Now, ab=\(​​\left( \frac { -3 }{ 2 } \right) \)(-2)=3
Hence, the value of ab is 3.

\(x = {4 \over 3}\) or 3

\(x^{2}+2x - 168 = 0\)

Yes, (x-1) (x+2) +2=0
\(\Rightarrow x^{ 2 }+x-2+2=0\)
\(\Rightarrow x^{ 2 }+x=0\)
\(\therefore \) b²-4ac=1>0
Distinct real roots

(i) 2x²+  \(5\sqrt { 3\quad } +6\quad =0\)
Here a =2, b  = \(5\sqrt { 3\quad } \) ,c=6
D=b²-4ac
\(\Rightarrow D=(5\sqrt { 3 } )^{ 2 }-4\times 2\times 6\)
=75-48=27 \(\Rightarrow D>0\)
\(\therefore \) Equation has real roots given by
\(x=\frac { -b+\sqrt { D } }{ 2a } ,\frac { -b-\sqrt { D } }{ 2a } \)
\(\Rightarrow x=\frac { -5\sqrt { 3 } +\sqrt { 27 } }{ 2\times 2 } ,\frac { -5\sqrt { 3 } -\sqrt { 27 } }{ 2\times 2 } \)
\(=\frac { -\sqrt { 3 } }{ 2 } ,-2\sqrt { 3 } \)

Let number of toys be x
\(\therefore \) Cost of one toy = \(\frac { 180 }{ x } \)
Number of toys sold = x-1
Selling price per toy = Rs \(\left( \frac { 180 }{ x } +1 \right) \)
Total SP =CP+profit
ATQ (x-1)  \(\left( \frac { 180 }{ x } +1 \right) \)=180+10
\(\Rightarrow (x-1)\left( \frac { 180 }{ x } +1 \right) =190\)
\(\Rightarrow x^{ 2 }+179x-180=190x\)
\(\Rightarrow x^{ 2 }-11x-180=0\)
\(\Rightarrow (x-20)(x+9)=0\)
\(\Rightarrow x=20,-9\)
\(\Rightarrow x=20\) [Rejected x=-9]

\(\frac { 2x }{ x-3 } +\frac { 1 }{ 2x+3 } +\frac { 3x+9 }{ (x-3)(2x+3) } =0\)
\(\Rightarrow \quad \frac { 2x(2x+3)+x-3+3x+9 }{ (x-3)(2x+3) } =0\)
\(\Rightarrow\) 4x²+6x+x-3+3x+9=0
\(\Rightarrow\) 4x²+10x+6=0
\(\Rightarrow\) 2x²+5x+3x+3=0
\(\Rightarrow\) 2x(x+1)-3(x+1)=0
\(\Rightarrow\) (x+1)(2x+3)=0
When x=a, given equation is not defned, \(\therefore\) x=-1

12abx²-(9a²-8b²)x-6ab=0 \(\Rightarrow\) 12abx²-9a²x+8b2x-6ab=0
\(\Rightarrow\) 3ax(4bx-3a)+2b(4bx-3a)=0 \(\Rightarrow\) (4bx-3a)(3ax+2b)=0
\(\Rightarrow\) 4bx-3a=0 or 3ax+2b=0 \(\Rightarrow\)  x=\(\frac { 3a }{ 4b } \) or x=\(\frac { -2b }{ 3a } \)

100x²-20x+1=0
100x2-10x-10x+1=0
10x(10x-1)-1(10x-1)=0
(10x-1)(10x-1)=0
10x-1=0 or 10x-1=0
\(x={1\over 10}\) or \(x={1\over 10}\)
Hence the roots are \({1\over 10},{1\over 10}\)

x²+5x-3=0
Degree of the equation is 2,
It is a quadratic equation.

The given equation is, (x²+1)(x+2)=(x+3)²
x³+2x²+x+2=x²+6x+9
x³+x²-5x-7=0
It is of degree 3 and is not of the form ax²+bx+c=0
Hence, given equation is not a quadratic equation.

Given equation (x-3)(2x+1)=x(x+5)
2x² + x - 6x - 3 = x² + 5x
\(\Rightarrow\) 2x² + x - 6x - 3 - x² - 5x = 0
\(\Rightarrow\) x² - 10x - 3 = 0 has its highest power 2
Hence it is a quadratic equation.

Let the usual speed of plane be x km/h,
\(\frac { 1500 }{ x } -\frac { 1500 }{ x+250 } =\frac { 1 }{ 2 } \)
\(\Rightarrow x^{ 2 }+250x-750000=0\)
\((x+1000)(x-750)=0\)
x = 750
Speed of plane = 750 km/h.
Values depicted here are:
(i) Helping the needy.
(ii) Quick help to the injured.

By question  \(T_{ passenger }=\frac { 192 }{ x } \)

\(\Rightarrow \frac { 192 }{ x } -\frac { 192 }{ x+16 } =2\)
\(\Rightarrow 192(x+16)-192x=2(x^{ 2 }+16x)\)
\(\Rightarrow 192x+192\times 16-192x=2(x^{ 2 }+16x)\)
\(\Rightarrow x^{ 2 }+48x-32x-1536=0\)
\(\Rightarrow x(x+48)-32(x+48)=0\)
\(\Rightarrow x(x-32(x+48)=0\)
\(\Rightarrow (x-32)(x+48)=0\)
x=32 or -48
Since speed can't be negative, therefore - 48 is not possible.
:. Speed of passenger train = 32 km/!h % and Speed of fast train = 48 km/h.

\(\sqrt { 3x^{ 2 } } -2\sqrt { 2 } x-2\sqrt { 3 } =0\)
\(\Rightarrow \sqrt { 3 } x^{ 2 }-3\sqrt { 2 } x+\sqrt { 2x } -2\sqrt { 3 } =0\)
\(\Rightarrow \sqrt { 3x } (x-\sqrt { 6 } )+\sqrt { 2 } (x-\sqrt { 6 } )=0\)
\(\Rightarrow (x-\sqrt { 6 } )(\sqrt { 3 } x+\sqrt { 2 } )=0\)
\(x=\sqrt { 6 } ,x=-\sqrt { \frac { 2 }{ 3 } } \)
Alternative method
\(\sqrt { 3x^{ 2 } } -2\sqrt { 2 } x-2\sqrt { 3 } =0\) 
we know that \(2\sqrt { 2 } \) can be written as \(3\sqrt { 2 } -\sqrt { 2 } \) 
\(\Rightarrow \sqrt { 3 } x^{ 2 }-\sqrt { 3\sqrt { 3 } } \sqrt { 2x } -\sqrt { 2 } \sqrt { 2 } \sqrt { 3 } =0\)
\(\Rightarrow \sqrt { 3 } x[x-\sqrt { 3 } \sqrt { 2 } ]+\sqrt { 2 } [x-\sqrt { 2 } \sqrt { 3 } =0\)
\(\Rightarrow \sqrt { 3 } x[x-\sqrt { 6 } ]+\sqrt { 2 } [x-\sqrt { 6 } ]=0\)
\(\Rightarrow x=\sqrt { 6 } orx=\frac { \sqrt { 2 } }{ \sqrt { 3 } } =-\sqrt { \frac { 2 }{ 3 } } \) 

\(4\sqrt { 3 } x^{ 2 }+5x-2\sqrt { 3 } =0\)
\(\Rightarrow 4\sqrt { 3 } x^{ 2 }+8x-3x-2\sqrt { 3 } =0\)
\(4x(\sqrt { 3 } x+2)-\sqrt { 3 } (\sqrt { 3 } (\sqrt { 3x+2)=0 } \)
\(\Rightarrow (\sqrt { 3 } x+2)(4x-\sqrt { 3 } )=0\)
\(\Rightarrow x=-\frac { 2 }{ \sqrt { 3 } } \frac { \sqrt { 3 } }{ 4 } \)

(i) Given, (x-1)(x+2)=(x-3)(x+1) ....(i)
Here, in LHS (x-1) and (x+2) are in product and in RHS (x-3) and (x+1) are also in product. So, first, we simplify their product.
LHS=(x-1)(x+2)=x²+2x-x-2=x²+x-2
RHS=(x-3)(x+1)=x²+x-3x-3=x²-2x-3
On substituting these values in Eq(i), we get
x²+x-2=x²-2x-3
x²-x²+x+2x-2+3=0
3x+1=0
It is not of the form ax²+bx+c=0, \(a\neq 0\). Since, here a=0 and it is an equation of degree 1.
Hence, the given equation does not represent a quadratic equation.
(ii) Given, (x+2)²=4(x+3) ...(ii)
Here, the term (x+2) has power 2. So, first expand it.
LHS=(x+2)²=x²+4+4x=4x+12
x²+4x-4x+4-12=0
x²-8=0 or x²+0x-8=0
It is of the form ax²+bx+c=0, \(a\neq 0\)
Hence, given equation represents a quadratic equation.

Given quadratic equation  is 
\((k-12)x^{ 2 }+2(k-12)x+2=0\)
On comparing with \(ax^{ 2 }+bx+x=0,\) we get
a=k-12,b=2(k-12) and c=2
Now, \(D=b^{ 2 }-4ac\)
=\([2(k-12)]^{ 2 }-4(k-12)(2)\)
\(=4(k-12)^{ 2 }-8(k-12)\)
=\((k-12)[4(k-12)-8]\)
= \((k-12)(4k-48-8)\)
=(k-12)(4k-56)

-5,-4 or 4,5

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

\(\frac { 4x }{ x-2 } -\frac { 3x }{ x-1 } =7\frac { 1 }{ 2 } \)
\(\Rightarrow \frac { 4x(x-1)-3x(x-2) }{ (x-2)(x-1) } =\frac { 15 }{ 2 } \)
\(\Rightarrow \frac { 4x^{ 2 }-4x-3x^{ 2 }+6x }{ x^{ 2 }-3x+2 } =\frac { 15 }{ 2 } \)
\(\Rightarrow 2(x^{ 2 }+2x)=15(x^{ 2 }-3x+2)\)
\(\Rightarrow 2x^{ 2 }+4x=15x^{ 2 }-45x+30\)
\(\Rightarrow 13x^{ 2 }+49x+30=0\)
\(\Rightarrow 13x-39x-10x+30=0\)
\(\Rightarrow (x-3)(13x-10)=0\)
\(\Rightarrow x=3\quad x=\frac { 10 }{ 13 } \)

Let speed of fast train be x km/h    \(\Rightarrow \) speed of slow train be(x-10)km/h
ATQ \(\frac { 600 }{ x-10 } -\frac { 600 }{ x } =3\)   \(\Rightarrow \)  \(\frac { 600x-600x+6000 }{ x(x-10) } =3\)
\(\Rightarrow \quad 6000=3x(x-10) \Rightarrow \ 2000={ x }^{ 2 }-10x\)
\(\Rightarrow x^{ 2 }-10x-2000=0\Rightarrow { x }^{ 2 }-50x+40x-2000=0\)
\(\Rightarrow (x-50)(x+40)=0\Rightarrow x=50,\quad -40\) (rejected)
Speed of fast train = 50 Km/h and speed of slow train = 50 - 10 =40 Km/h

Since (-5) is a root of siven quadratic equation  2x² + px - 15= 0and the qudratic equation P(x² + x) + k = 0 has equal roots, then find the values of p and k
2(-5)²+p(-5)-15=0
50+5p-15=0
5p=35p=7
Now p(x²+x)+k =0 has equal roots
Px²+px+k=0
So (b)^2_--4ac=0
(p)²-4p\(p\times k=0\)
(7)²-4x7xk=0
28k=49
\(k=\frac { 49 }{ 28 } =\frac { 7 }{ 4 } \)
p=7 k=\(\frac { 7 }{ 4 } \)

If x = - 4 is a root of the equation x²+ 2x + 4p = 0
p=-2
Equation  x² - 2 (1+ 3k) x + 7 (3+ 2k) = 0 has equal roots.
4(1 + 3k)² - 28(3 + 2k) = 0
\(\Rightarrow 9k^{ 2 }-8k-20=0\)
\(\Rightarrow (9k+10)(k-2)=0\)
\(\Rightarrow k=\frac { -10 }{ 9 } ,2\)

Let marks in Mathematics = x
Therefore, marks in Science ::; 32 - x
\(\Rightarrow (32-x-2)(x+4)=253\)
\(\Rightarrow (30-x)(x+4)=253\)
\(\Rightarrow 26x-x^{ 2 }+120=253\)
\(\Rightarrow x^{ 2 }-26x+133=0\)
\(\Rightarrow x^{ 2 }-19x-7x+133=0\)
\(x(x-19)-7(x-19)=0\)
x=7 or x=19
If x = 7 \hen marks in Mathematics =7 
marks in Science = 25
and if x = 19, then
marks in Mathematics = 19 
marks in Science = 13.

12,16

Let speed of the stream = x km/h
Given, speed of boat in still water = 15 km/h
Speed of boat upstream = (15 - x)km/h
and speed of boat downstream = (15 + x) km/h
According to the question,
\(\frac { 30(15+x)+30(15-x) }{ (15-x)(15+x) } =\frac { 9 }{ 2 } \)
\( \frac { 450+30x+450-30x }{ { \left( 15 \right) }^{ 2 }-{ x }^{ 2 } } =\frac { 9 }{ 2 } \)
\(\frac { 900 }{ 225-{ x }^{ 2 } } =\frac { 9 }{ 2 }\)
\(\frac { 900\times 2 }{ 9 } =225-{ x }^{ 2 }\)
\(200=225-{ x }^{ 2 }\)
\({ x }^{ 2 }-225+200=0\)
\({ x }^{ 2 }-25=0\)
\({ x }^{ 2 }=25\)
\(x=\pm 5\)
But speed cannot be negative.
x = 5
Hence speed of stream is 5km/h.

45 km/h.

Given equation is \(2x^{ 2 }-x\frac { 1 }{ 8 } =0\)
On multiplying both sides by 8, we get  \(16x^{ 2 }-8x+1=0\)
\(\Rightarrow 16x^{ 2 }-4x-4x+1=0\)
\(\left[ \because (-4)X(-4)=16\quad and\quad -4-4=-8 \right] \)
\(\Rightarrow 4x(4x-1)-1(4x-1)=0\)
\(\Rightarrow 4x-1=0\Rightarrow x=\frac { 1 }{ 4 } \)
and 4x-1=0  \(\Rightarrow x=\frac { 1 }{ 4 } \)
Hence the roots of the equation \(2x^{ 2 }-x+\frac { 1 }{ 8 } =0\)are \(\frac { 1 }{ 4 } \quad and\quad \frac { 1 }{ 4 } \)

Given equation is,  (x-3)(2x+1)=x(x+5)
\(\Rightarrow 2x^{ 2 }-6x+x-3=x^{ 2 }+5x\)
\(\Rightarrow 2x^{ 2 }-6x+x-3=x^{ 2 }+5x=0\)
\(\Rightarrow x^{ 2 }-10x-3=0\)
Which is of the form \(ax^{ 2 }+bx+c=0\) where and b,c are any real numbers.Hence,it is a quadratic equation.

Let Integers be x,x+1 and x+2
ATQ x(x+1) (x+2) =16(x+x+1+x+2)
\(\Rightarrow (x^{ 2 }+x)(x+2)=16(3x+3)\)
\(\Rightarrow x^{ 3 }+2x^{ 2 }+x^{ 2 }+2x=48+48 \)
\(\Rightarrow x^{ 3 }+3x^{ 2 }-46x-48=0\)
LHS when x=-1, we have
(x+1) is a factor
For others factor

\(\therefore x^{ 3 }+3x^{ 2 }-46x-48=0\)
\(\Rightarrow (x+1)(x^{ 2 }+2x-48=0)\)
\(\Rightarrow (x+1)(x+8)(x-6)=0\)
\(\Rightarrow x=-1,x=-8,x=6\)
Rejecting -ve values x = 6 \(\therefore \) positive integers are 6,