In each of these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer.

I. $4x^2-16x+16=0$
II. $10y^2-29y+21=0$

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The correct answer is If x > y.

To determine the relationship between $x$ and $y$, let us solve both quadratic equations step-by-step.

Step 1: Solve Equation (I) for $x$:

$4x^2-16x+16=0$

Divide the entire equation by 4 to simplify:

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

This is a perfect square trinomial:

${(x-2)}^2=0$

$x=2$

So, the value of $x$ is $2$.

Step 2: Solve Equation (II) for $y$:

$10y^2-29y+21=0$

We need two numbers that multiply to $10\times 21=210$ and add up to $-29$. These numbers are $-14$ and $-15$.

Split the middle term:

$10y^2-15y-14y+21=0$

Factor by grouping:

$5y(2y-3)-7(2y-3)=0$

$(2y-3)(5y-7)=0$

Thus, the roots are:

$y=\frac{3}{2}=1.5$

or

$y=\frac{7}{5}=1.4$

Step 3: Compare the values of $x$ and $y$:

Value of $x=2$

Values of $y=1.5,1.4$

Comparing $x$ with both values of $y$:

• $2>1.5$

• $2>1.4$

In both cases, $x>y$.

Therefore, the correct relationship is If x > y.