Directions: In the following question, two equations numbered I and II are given. Solve both equations and determine the relationship between x and y.

I. $2x^2+13x+20=0$
II. $2y^2+7y+6=0$

The correct option is If x < y.

Step 1: Solve Equation I for x

The first quadratic equation is:

$2x^2+13x+20=0$

Factorize the quadratic expression by splitting the middle term:

$2x^2+8x+5x+20=0$

$2x(x+4)+5(x+4)=0$

$(2x+5)(x+4)=0$

Set each factor equal to zero to find the roots:

$2x+5=0\Rightarrow x=-\frac{5}{2}=-2.5$

$x+4=0\Rightarrow x=-4$

Thus, the values of x are $x=-2.5$ and $x=-4$.

Step 2: Solve Equation II for y

The second quadratic equation is:

$2y^2+7y+6=0$

Factorize the quadratic expression by splitting the middle term:

$2y^2+4y+3y+6=0$

$2y(y+2)+3(y+2)=0$

$(2y+3)(y+2)=0$

Set each factor equal to zero to find the roots:

$2y+3=0\Rightarrow y=-\frac{3}{2}=-1.5$

$y+2=0\Rightarrow y=-2$

Thus, the values of y are $y=-1.5$ and $y=-2$.

Step 3: Compare the values of x and y

- Comparing $x=-2.5$ with $y=-1.5$: $-2.5<-1.5\Rightarrow x<y$
- Comparing $x=-2.5$ with $y=-2$: $-2.5<-2\Rightarrow x<y$
- Comparing $x=-4$ with $y=-1.5$: $-4<-1.5\Rightarrow x<y$
- Comparing $x=-4$ with $y=-2$: $-4<-2\Rightarrow x<y$

Since every value of x is strictly smaller than every value of y, the relation between x and y is $x<y$.