Given the quadratic equations:

I. $(4x^2 - 16x + 16 = 0)$

II. $(10y^2 - 29y + 21 = 0)$

Determine the correct relationship between $x$ and $y$ where $x$ and $y$ are the roots of the respective equations.

Question

Given the quadratic equations:

I. $(4x^2 - 16x + 16 = 0)$

II. $(10y^2 - 29y + 21 = 0)$

Determine the correct relationship between $x$ and $y$ where $x$ and $y$ are the roots of the respective equations.

Accepted Answer

**Solution:**

1. Consider the first equation:
   $$4x^2 - 16x + 16 = 0$$
   We can rewrite it as:
   $$4x^2 - 8x - 8x + 16 = 0$$

2. Group terms:
   $$4x(x - 2) - 8(x - 2) = 0$$

3. Factor out common terms:
   $$(4x - 8)(x - 2) = 0$$

4. Set each factor to zero:
   $$4x - 8 = 0 \Rightarrow x = 2$$
   $$x - 2 = 0 \Rightarrow x = 2$$

So, the roots of the first equation are both $x = 2$.

5. Now, consider the second equation:
   $$10y^2 - 29y + 21 = 0$$
   Rewrite as:
   $$10y^2 - 15y - 14y + 21 = 0$$

6. Group terms:
   $$5y(2y - 3) - 7(2y - 3) = 0$$

7. Factor common terms:
   $$(5y - 7)(2y - 3) = 0$$

8. Set each factor to zero:
   $$5y - 7 = 0 \Rightarrow y = \frac{7}{5} = 1.4$$
   $$2y - 3 = 0 \Rightarrow y = \frac{3}{2} = 1.5$$

9. The roots of the second equation are $y = 1.4$ and $y = 1.5$.

10. Comparing the roots:
    - $x = 2$
    - $y = 1.4$ or $1.5$

Since $2 > 1.5$, we conclude:

**Answer:** $x > y$

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