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

I. $x^2-9x+20=0$
II. $2y^2-7y+6=0$

Correct Answer: If x > y

To determine the relation between $x$ and $y$, we solve both quadratic equations step-by-step and compare their roots.

Step 1: Solve Equation I for $x$

The first equation is:

$x^2-9x+20=0$

We factorize the quadratic equation by splitting the middle term into two numbers whose sum is -9 and product is 20. These numbers are -4 and -5.

$x^2-4x-5x+20=0$

$x(x-4)-5(x-4)=0$

$(x-4)(x-5)=0$

Equating each factor to zero:

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

$x-5=0\Rightarrow x=5$

So, the roots for $x$ are $4$ and $5$.

Step 2: Solve Equation II for $y$

The second equation is:

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

We split the middle term into two numbers whose sum is -7 and product is $2\times 6=12$. These numbers are -4 and -3.

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

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

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

Equating each factor to zero:

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

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

So, the roots for $y$ are $1.5$ and $2$.

Step 3: Comparison of $x$ and $y$ values

Comparing every root of $x$ with every root of $y$:

• For $x=4$ and $y=1.5$: $4>1.5$
• For $x=4$ and $y=2$: $4>2$
• For $x=5$ and $y=1.5$: $5>1.5$
• For $x=5$ and $y=2$: $5>2$

Since every value of $x$ is strictly greater than every value of $y$, the correct relationship is x > y.