Solving Quadratic Equation I: Finding Values for x

We need to solve the first equation: $x^2 – 26x + 165 = 0$. This is a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a=1$, $b=-26$, and $c=165$. We can find the values of x by factoring the equation.

We are looking for two numbers that multiply to 165 ($c$) and add up to -26 ($b$). Let's list the factors of 165:

Since we need the sum to be -26, the two numbers must be -11 and -15 (because $(-11) \times (-15) = 165$ and $(-11) + (-15) = -26$).

Now, we can rewrite the equation:

$(x - 11)(x - 15) = 0$

For this equation to be true, either $(x - 11) = 0$ or $(x - 15) = 0$.

• If $x - 11 = 0$, then $x = 11$.
• If $x - 15 = 0$, then $x = 15$.

So, the possible values for x are 11 and 15.

Solving Quadratic Equation II: Finding Values for y

Next, we solve the second equation: $y^2 – 38y + 357 = 0$. This is also a quadratic equation where $a=1$, $b=-38$, and $c=357$. We will find the values of y by factoring.

We need two numbers that multiply to 357 ($c$) and add up to -38 ($b$). Let's find the factors of 357:

To get a sum of -38, we need the numbers -17 and -21 (because $(-17) \times (-21) = 357$ and $(-17) + (-21) = -38$).

Rewrite the equation using these factors:

$(y - 17)(y - 21) = 0$

This implies either $(y - 17) = 0$ or $(y - 21) = 0$.

• If $y - 17 = 0$, then $y = 17$.
• If $y - 21 = 0$, then $y = 21$.

Therefore, the possible values for y are 17 and 21.

Comparing x and y Values

Now we compare the values obtained for x and y. The values for x are {11, 15}. The values for y are {17, 21}.

Let's compare each possible value of x with each possible value of y:

• Compare $x=11$ with $y=17$: $11 < 17$.
• Compare $x=11$ with $y=21$: $11 < 21$.
• Compare $x=15$ with $y=17$: $15 < 17$.
• Compare $x=15$ with $y=21$: $15 < 21$.

In all possible comparisons, every value of x is less than every value of y. Therefore, the relationship is $x < y$.