\(\sqrt { 3x^{ 2 } } -2x-\sqrt { 3 } =0\)
⇒ \(\sqrt { 3x^{ 2 } } -3x+x-\sqrt { 3 } =0\)
⇒ \(\sqrt { 3 } x(x-\sqrt { 3 } )+1(x-\sqrt { 3 } )=0\)
⇒ \((x-\sqrt { 3 } )(\sqrt { 3 } +1)=0\)
∴ x =\(\sqrt { 3 },\frac{-1}{\sqrt { 3 }}\)

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 = \(\frac{-9}{4}\)

For equal roots, we have
b²-4ac=0
\(\Rightarrow\)  (k+1)²-4(k+4)(1)=0
\(\Rightarrow\) k²+2k+1-4k-16=0
\(\Rightarrow\) k²-2k-15=0
\(\Rightarrow\) (k-5)(k+3)=0
\(\Rightarrow\)  k=5 or k=-3

Since x-2 is a root of given equation
\(\therefore\) (2)²+2k(2)+4=0
\(\Rightarrow\) 4+4k+4=0
\(\Rightarrow\) 4k=-8
\(\Rightarrow\) k=-2
Hence, the value of k is -2.

As 8 is a root of the equation x²-10x+k=0
∴ (8)²-10(8)+k=0
⇒ 64-80+k=0 ⇒ k=16
Hence, the value of k is 16.

K = -14

\((\sqrt {3} \ x \ - \sqrt {2} ) (\sqrt {3} \ x \ -\sqrt {2} )\)

Given equation is kx²+2x+k=0
Here, a=k,b= 2,c=k
For equal roots,D=0
\(\Rightarrow b^{ 2 }-4ac=0\)
(2)² - 4 x k x k=0
\(\Rightarrow 4k^{ 2 }=4\Rightarrow k^{ 2 }=1\Rightarrow 1k=\pm 1\)

For equal roots, D = 0

Here a= k , b=4 ,c=1
D=b²-4ac = 4²-4x k x 1=16-4k
For equal roots,D=0
\(\Rightarrow 16-4k=0\Rightarrow k=4\)

For unequal and real roots, D > 0

Given quadratic equation is 4x²+6x+3=0
Here, a=4, b=6, c=3
D=b²-4ac
D=(6)²-4x4x3=36-48=-12< 0
given quadratic equation has no real roots.

Given quadratic equation  9x²-6x-2=0
Here a=9, b=-6, c=-2
D=b²-4ac
D=(-6)²-4X9X(-2)=36+72=108 > 0
Given quadratic has two unequal real roots.

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]

Let number of days taken by B to finish the work be x number of days taken by A to finish the work = (x-10) Number of days taken by A and B to finish the work = 12 Now, A's one day work + B's one day work 
= one day's work A and B
\(\Rightarrow \frac { 1 }{ x-10 } +\frac { 1 }{ x } =\frac { 1 }{ 12 } \)
\(\Rightarrow \frac { x+x-10 }{ (x-10)x } =\frac { 1 }{ 12 } \)
\(\Rightarrow 24x-120={ x }^{ 2 }-10\)
\(\Rightarrow { x }^{ 2 }-34x+120=0\)
\(\Rightarrow \left( x-30 \right) \left( x-4 \right) =0\)
x =30, x= 4 (not possible)
Number of days taken by B = 30

x²-2x-8=0
When x=-2, LHS=(-2)²-2(-2)+8=4+4+8=16≠0
x=-2 is not a solution of the given equation.

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=\(\frac { 7 }{ 4 } \)

\(\frac { x-3 }{ x-4 } +\frac { x-5 }{ x-6 } =\frac { 10 }{ 3 } \)
\(\frac { (x-3)(x-6)+(x-4)(x-5) }{ (x-4)(x-6) } =\frac { 10 }{ 3 } \)
\(\frac { (x-3)(x-6)+(x-4)(x-5) }{ (x-4)(x-6) } =\frac { 10 }{ 3 } \)
\(3(2x^{ 2 }-18x+38)=10x^{ 2 }-100x+240\)
\(6x^{ 2 }-54x+114=10x^{ 2 }-100x+240\)
4x²-46x+126=0
2x²-14x-9x+63=0
2x(x-7)-9(x-7)=0
(2x-9)(x-7)=0
2x-9=0,x-7=0
x=9/2 , x=7

Let the speed of the stream be x km/h.
Therefore, the speed of the boat upstream = (18 - x) km/h and the speed of the boat downstream = (18 + x) km/h.
The time taken to go upstream = \(\frac{\text { distance }}{\text { speed }}=\frac{24}{18-x}\) hours.
Similarly, the time taken to go downstream = \(\frac{24}{18+x}\) hours.
According to the question,
\(\frac{24}{18-x}-\frac{24}{18+x}=1\)
i.e., 24(18 + x) - 24(18 - x) = (18 - x) (18 + x)
i.e., x² + 48x - 324 = 0
Using the quadratic formula, we get
\(x =\frac{-48 \pm \sqrt{48^{2}+1296}}{2}=\frac{-48 \pm \sqrt{3600}}{2} \)
\(=\frac{-48 \pm 60}{2}=6 \text { or }-54 \)
Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = - 54. Therefore, x = 6 gives the speed of the stream as 6 km/h.