Explanation :

Let α and β be the roots of the general form of the quadratic equation: ax² + bx + c = 0.

Here, α = 2 and β = $\dfrac{-1}{2}$

Sum of the roots = α + β

⇒ α + β = 2 + $\left(\dfrac{-1}{2}\right)$ = $\dfrac{3}{2}$

Product of the roots = αβ

⇒ αβ = 2 x $\left(\dfrac{-1}{2}\right)$ = - 1

Therefore, the required equation is:

    x² - (α + β)x + αβ = 0

⇒ x² - $\dfrac{3}{2}$ x - 1 = 0

⇒ 2x² - 3x - 2 = 0

Explanation :

∵ 3x² + 2x -1 = 0

⇒ 3x(x + 1) - 1(x + 1) = 0

⇒ (3x - 1)(x + 1) = 0

⇒ x = -1, $\dfrac{1}{3}$

∵ 2y² + 7y + 6 = 0

⇒ 2y² + 4y + 3y + 6 = 0

⇒ 2y(y + 2) + 3(y + 2) = 0

⇒ (2y + 3)(y + 2) = 0

⇒ y = -2, $\dfrac{-3}{2} $

For x = $\dfrac{1}{3}$, x > y for both values of y i.e. -2, $\dfrac{-3}{2} $

and, For x = -1, x > y for both values of y i.e. -2, $\dfrac{-3}{2} $

Hence, for above result we get that x > y.

Explanation :

$7x^2 + 8x + 1 = 0 \\ \Rightarrow 7x^2 + 7x +x+ 1 = 0 \\ \Rightarrow 7x(x+1)+1(x+1)=0 \\ \Rightarrow (7x+1)(x+1)=0 \\ \Rightarrow x=-1,\cfrac{-1}{7} \\ Now, 18y^2 + 11y = -1 \\ \Rightarrow 18y^2 + 9y +2y + 1=0 \\ \Rightarrow 9y(2y+1)+1(2y+1)=0 \\ \Rightarrow (9y+1)(2y+1)=0 \\ \Rightarrow y=\cfrac{-1}{2},\cfrac{-1}{9}$

For x=-1,x>y for both values of y i.e -1/2 and -1/9 

and, For x=-1/7, when y=-1/2 then x > y and when y= -1/9 then x< y

For above two conclusion there is no relationship x and y

Explanation :

$x= \cfrac{\sqrt{16}}{\sqrt{36}} \\ \Rightarrow x= \cfrac{4}{9} \\ \Rightarrow x= \cfrac{2}{3} \\Now,  9y^2 – 39y + 42 = 0 \\ \Rightarrow  9y^2 – 18y-21y + 42 = 0 \\ \Rightarrow 9y(y-2)-21(y-2)=0 \\ \Rightarrow (9y-21)(y-2)=0 \\ \Rightarrow y=\cfrac{21}{9},2 \\ \Rightarrow y=\cfrac{7}{3},2$

x < y for all values of y.