For each quadratic equation, let x and y be any real roots, respectively. Select the most precise relationship between x and y.

$\begin{array}{l}\text{I.}{x}^2+10x+21=0\\ \text{II.}{y}^2-5y+4=0\end{array}$

Correct Answer: If x < y

To determine the most precise relationship between the real roots x and y, we solve each quadratic equation individually and then compare their roots.

Step 1: Solve Equation I to find the values of x

The first equation is given as:

$x^2+10x+21=0$

We factorize the quadratic equation by finding two numbers whose product is 21 and sum is 10. These numbers are 7 and 3:

$(x+7)(x+3)=0$

Setting each factor equal to zero gives:

$x+7=0\text{⇒}x=-7$

$x+3=0\text{⇒}x=-3$

Thus, the roots for x are x = -7 and x = -3.

Step 2: Solve Equation II to find the values of y

The second equation is given as:

$y^2-5y+4=0$

We factorize this quadratic equation by finding two numbers whose product is 4 and sum is -5. These numbers are -4 and -1:

$(y-4)(y-1)=0$

Setting each factor equal to zero gives:

$y-4=0\text{⇒}y=4$

$y-1=0\text{⇒}y=1$

Thus, the roots for y are y = 4 and y = 1.

Step 3: Compare the values of x and y

Now, we compare each possible root of x with each possible root of y:
- For x = -7 and y = 4: -7 < 4
- For x = -7 and y = 1: -7 < 1
- For x = -3 and y = 4: -3 < 4
- For x = -3 and y = 1: -3 < 1

In all cases, every possible value of x is strictly less than every possible value of y.

Therefore, the correct relationship between x and y is If x < y.